Killer Math Problem With An Unbelievably Elegant Solution

[@dhasarathisrikrishna2335]

I can’t believe how creatively this question was solved!

[@TicoTimeCR]

Looks like many people (like myself) had an issue with the wording of the question. Perhaps this would be better? : "If the line segments from each vertex to the random point were rearranged to form a new triangle, what would the new triangle's angles be?"

[@SeTConVar]

agreed.

[@agfd5659]

It may be better wording, but the original is not confusing at all. What else could it have meant?

[@Lifylives]

Not totally agreed brother

[@TheKivifreak]

Well given 3 segments there is either zero or one possible triangle that can be made out of the segments (up to symmetry).

[@jumpingjflash]

I had to watch the video to find out what a line segment is.

[@ColinRichardson]

Did I figure it out??? I still can't figure out what the question is....

[@grantchristopher170]

Take the segments created by drawing to the random point from the vertices and rearrange them to create a new triangle. What are the angles of that new triangle?

[@Czeckie]

yep, the original video by GoldPlatedGoof explains it well and is overall more engaging.

[@think4all]

U need meditation

[@trekkingandcycling4889]

@@think4all hahaha

[@simon_patterson]

Ha, same here!

[@georgewang2947]

I actually figured it out! It took me several days of drawing and trying various methods before I got it ☺️

[@xXDarQXx]

It took me 5 min LMAO. Although I might have had a head start after watching this video: kzbin.info/www/bejne/gYKblIarpNSjr8U.

[@sakshamtyagi6527]

It took me an eternity to understand what the question was saying ...

[@Sl-by1ss]

That’s awesome man. Congrats.

[@nagualdesign]

Bravo! It goes to show that the hardest part of any of these problems is resisting the urge to simply watch the rest of the video without actually trying to figure out the solution.

[@RustyBrakes]

It's actually nice to hear that Presh can't solve some of these at first - it's a bit like when you watch someone's highlight reel and don't see all the failed attempts. thanks for including those details!

[@TheHuesSciTech]

I solved this using complex numbers. There was still "a trick" to it, but I think the trick was much more self-evident in this approach. It goes like this: I choose the vertices of my triangle to be describe as points on the complex plane, specifically the three cube roots of 1. So they are A, which is equal to 1, and B and C, which are the other two complex roots. Since these are all roots of unity, all sorts of identites fall out naturally: A+B+C=0, B*C=A, etc etc. The random centre point is called p. So the task is to find a way to take the three edge vectors (A-p), (B-p), (C-p), and form them into a triangle. If they're in a triangle, the three edge vectors will add to zero. So we could try just naively adding them together, and we get (A-p) + (B-p) + (C-p) = 3p, which isn't zero (except for the special case p=0), so we have to rotate the edges somehow to get them to fit in a triangle. So we need to multiply (i.e. rotate) those vectors by some magnitude-1 numbers so that the p's cancel out. It only takes a quick bit of experimentation to find that A*(A-p) + B*(B-p) + C*(C-p) = A+B+C - (A+B+C)*p = 0 - 0*p = 0 works. So the edge going to A is left untouched, and the other two are rotated 120 degrees in different directions. The edges then slot together perfectly into a triangle, and we know exactly how much we rotated them by so we can draw a diagram and infer the internal angles. I can't draw that diagram here, but it works out that you add 120 degrees for the rotation, but take away 180 degrees to get the internal rather than the external angle, and so you take away (180-120)=60 degrees in total.

[@NiGHTM4RE714]

Does anyone else also try to figure it out by looking at the thumbnail and if you can't, you decide to watch the video??

[@noahalexander2019]

Meeee

[@laxmiparida4119]

Meeeeeeeeeeeeeee

[@peaceistherealmuscle]

I’ll show this problem to my math teacher tomorrow. He loves solving puzzles like this one

[@andrewgonzales2284]

Mr. Cool Guy did he solve it?

[@peaceistherealmuscle]

Yep, handed me the solution the day after :D

[@joshuamason2227]

maybe searched online lol

[@karthikeyangovindaraj7932]

Wu Chenglin Savage

[@jakeroberts6102]

Maths*

[@NotYourAverageNothing]

I was stuck on the wording of the problem…

[@jabir5768]

IKR

[@yaskynemma9220]

The problem is understanding the question

[@skylardeslypere9909]

I wasn't and i'm not even native english

[@yaskynemma9220]

Skylar Deslypere well, i am mexican but i have to tell i admire you

[@agfd5659]

The wording seems quite straight forward and easy to me, I don't see, what you could have had a problem with.

[@MartinPuskin]

What a coincidence, just today I looked at that video about the "coffin" problems and tried my hand at some of them! I actually managed to solve two of them (both on algebra). Really beautiful and well-presented solution on this one, Presh!

[@codebeard]

I was able to prove this without any prior knowledge of tricks or information about how to solve the problem. First I looked at a few special cases for where the point could be (1. the exact centre of the triangle, 2. the midpoint of one of the edges, 3. one of the vertices) and noticed that in each case the angles in the triangle were always 60 degrees less than corresponding angles at the vertex. Working with this as a hypothesis for what would happen to any point inside the triangle, I realised that an easy way to subtract 60 degrees from an angle was to construct an equilateral triangle at the point using one of the lines to a vertex. In the end I came up with a similar construction to your diagram but without rotating the whole triangle, effectively just rotating two edges. I enjoyed the problem a lot, thank you for sharing.

[@ordinaryperson883]

codebeard well you kinds got lucky there no offense someone can take is as half of the original Triangle

[@carlfranz6805]

You have my admiration. Good on you.

[@sidharthr6615]

Archana Sharma that's exactly what I did ...lol...I got x/2, y/2 as the two angles

[@arcchitjain]

I solved (but not proved) it exactly like this too.

[@arcchitjain]

the solution corresponding to the third case (i.e. when the random point is shifted to one of the vertices) would not satisfy if you would consider half of the original angles. This is because the original angles, (60.2, x = 149.9, y = 149.9) would transform to (0.2, x = 89.9, y = 89.9).

[@srinibasdas9123]

Please consider this alternative approach. Let us name the lines from vertices A,B & C to the arbitrary point O as a,b & c. Now draw a triangle with sides a' ,b' & c' such that a & a' ,b&b' and c& c' are parallel pairs. This is just like Vector polygon to present a FORCE system in Physics.. Now 2 angles of the triangle with sides a',b' and c' are 180-x & 180- y. So the third angle is x+y-180.

[@Arsiana]

I don't understand why are people so complaining about "wording". I do not speak English well, so sorry for bad grammar, but I was still able to figure it out (the question, not the answer). Maybe a little bit of thinking before smashing that comment would make understanding more possible. And to the video: as I'm a math student, I really appreciate this type of videos. It teaches me how to think "outside the box" and to think about problems more deeply, so thanks a lot for that! Your videos gave me more, than some of my professors at university.

[@leif1075]

@evisceron it is a little vague in terms of how to construct the new triangle ...you take the three sides..but how do you arrange them..but I see since they are fixed line segments theres only one way to male them all,connect into a closed polygon..

[@janetim5452]

agree, they are either refusing to think or thinking too much...

[@zoetropo1]

It's because the word 'remove' has multiple, incompatible meanings.

[@shreeganesh441]

I did this problem 2 years ago, but in a different way. The proof you've given is awesome

[@defunct6613]

would like to see ur soln

[@bowlchamps37]

It is from a Russian regional math master test. Grade 11-12.

[@venugopalamn9307]

A wonderful job indeed, to inspire learning and teaching mathematics, with clarity. Kudos to you.

[@mohammed90158]

لا أعرف كيف أعبر عن شكري لك على المرئيات الرائعة التي تقدمها. والمسألة التي عرضتها في هذه المرئية بالذات كانت رائعة بحقّ.

[@davidgatzen1543]

I was able to figure it out. It took me a few hours. I started by trying to calculate everything I could. My break through came when I decided to make 3 copies of the original drawing and then cut out the 3 original equilateral triangles. This allowed me to move the triangles around like he did in the video. My solution was much more complicated than the solution in the video, but I was able to figure it out without having to use trigonometry.

[@shaswatadutta4451]

Truly speaking I didn't take even 2 minutes to do it.. but I appreciate the geometry.. hats off to the one who created it..

[@mohitanjana1000]

Bro i don't know but how do you that?? Share you solution through your Instagram story..

[@shaswatadutta4451]

@@mohitanjana1000 it's simple rotation of one of the triangles. Try to figure out which one.

[@azzamlabib4785]

Ah...the satisfied feeling when you see new geometry challenge...

[@EighteenCharacters]

I really, REALLY like this one. You have so many unknowns but you force a manipulation to create knowns from the unknowns to work it out. It's like saying there's 300 people who need to get to the airport, and 1 car... It's impossible... so lets turn all the people into cats using magic. Right? Knowing that rotating the triangle will create complimentary angles and you can start with something you DO know (complimentary angles) takes a fair bit of reasoning, and is quite elegant.

[@agfd5659]

@Fester Blats the value of x is known, though

[@abdixsimplix2582]

Yeah, but real life doesn't work like that. Those are problems who are created by knowing that there is a solution.

[@ddebenedictis]

Do cats know where the airport is though?

[@dlevi67]

Best one in a long while! Thank you for making the video, Presh.

[@curtellingboe2949]

A point at the center would result in an equilateral triangle with lengths equal to two-thirds the original. If the point is moved towards the opposite vertex, x would approach 60 degrees but form a triangle with a vanishing length opposite an angle approaching zero. If the point is moved towards the midpoint of the opposite side, x would approach 180 degrees but form a flat triangle with an angle approaching 120 degrees. All three cases show the corresponding angle is x - 60. And y - 60 is the analogous solution for the second angle.

[@CraigNull]

Another fairly simple proof method; my verbosity is for clarity more than complexity. Label the three line segments we're interested in as a, c, and e and label the vertices of the original triangle as A, C, E with A corresponding to segment a, etc. -- the weird labeling will be clearer in a moment. Take the three obtuse interior triangles with angles x, y, and 360-x-y and reflect each of them across its corresponding side of the equilateral triangle. This results in an irregular hexagon. Label the new vertices B, D, F, such that B is between A and C, etc. Examine triangle FAB. It's not hard to see angle FAB is 120 degrees, since it's effectively doubling the angle of the original triangle's vertex. Also triangle FAB is isosceles, with two of its sides having unfolded from the same interior line segment of length a, so we can infer angles ABF and AFB are each 30 degrees. Label side FB as having length k*a for some proportionality constant k. Likewise, triangles BCD and DEF are also isosceles with a 120 degree angle, so BD is length k*c and DF is length k*e, and they also introduce a bunch of new 30 degree angles stemming from vertices B, D, F. Now, triangle BDF is similar to our desired triangle, with side lengths k*a, k*c, k*e. Since these are the corners that reflected outward the full angle at each of these vertices are x, y, and 360-x-y, but the subangles associated with the triangle BDF are each of these diminished by two 30 degree angles. Thus they're x-60, y-60, 300-x-y.

[@JianJiaHe]

Yes, I have figured it out and am using the same approach. I think rotating object is actually a cliché in Mathematics... We see it everywhere, some versions of proof of Pythagorean theorem, proof of Fermat point, atan(1/2) + atan(1/3) = pi/4 proof, etc.

[@shigangshan7785]

Rotation is so important in solving this problem. I can use this method later. Thank you very much!

[@louf7178]

In engineering, I recall two frequent methods in solving derivations: adding 0 to an equation or multiplying by 1 ... conjuring what to introduce is the genius part; this seems similar in geometry with these approaches.

[@sarthu123]

The solution just blew my mind! Hats off to ingeniousness of whosoever discovered it on himself.

[@fiziwig]

That's really thinking outside the three-sided box. I can't imagine that I would have ever stumbled on that clever approach.

[@disgruntledtoons]

I looked at the limit situations when the "random" point approaches one of the corners, the mid-point of a side, and from the exact center. I observed that the angles of the resulting triangle differed from the angles between the lines by sixty degrees.

[@Quintaner]

I got really caught up in the wording of the problem. I’ve never heard of defining a triangle with the line segments like that, so I didn’t even know what was being asked

[@electronicmusicwave7459]

beautiful - i solved algebraicly call the vertex of the trianlge with equal sides A B C. A B and y define a cirlce. A C and x define a circle. Calculate Intersection Point (S). Calculate blue, yellow, violet lines (AS, BS, CS). Calculate angle by tan(alpha/2) = sqrt((u-b)(u-c)/(u(u-a)) done

[@willbishop1355]

A lot easier than trying to do it with the Law of Cosines!

[@stevethecatcouch6532]

I tried using the law of cosines, but quickly realized that was not the clever trick he alluded to.

[@assalane]

I tried to do it using linear algebra. Didn't go well

[@ordinaryperson883]

Law cosines took nearly 1 page to solve and moreover its not elegant

[@subhankarpramanik2224]

Will Bishop I was also trying to get it from the cosine rule...

[@1234vedas]

I tried it,it doesn't work!

[@charlesdalton9810]

I found the answer in my head in like 2 minutes. It helps if you think of specific cases like if the point were in the center, if it were one of the vertices, or if the point were one of the side’s midpoints.

[@stevethecatcouch6532]

Using some standard math puzzle meta analysis, I get the new triangle's angles as x-60, y-60 and 300-x-y, but the process is very unsatisfying. I'll probably have to sleep on this to come up with something with more meat on it.

[@sharonminsuk]

I would be interested to see that "standard" meta analysis spelled out.

[@anthonygambong3027]

If you define the angle z such that x+y+z=360 from the intersections, then if you rotate 60° from the vertex on both sides, then you have two equivalent triangles with the colored sides. I still got the same answer but my last angle was z-60 which is just equal to 300-x-y. Might be more complicated, but there are people that dont like two variables in an equation (like me)

[@ivanabcdefg9375]

Much easier solution: 1. Re-draw the diagram such that, for each black line of the equilateral triangle, you "mirror" the 2 adjacent colored inner lines over it. Now you have a hexagon whose every other angle is 120 degrees and the remaining angles are x, y, and z. (By the way, z is 360-x-y so it's basically a given). 2. With the hexagon, you can "flatten away" the 120 degree angles into 180 degree angles, forming a triangle whose sides are 2x, 2y, and 2z, and this is OK because the similar triangle to x,y,z will have equal angles. To flatten away, notice that you're subtracting 30 degrees from the adjacent angles, thus you're subtracting 60 degrees total from the 3 remaining angles. And that's your result: the angles of the triangle are 60 degrees less than their original values. X is x - 60; Y is y - 60; and Z is z - 60 (given in the video as 300 - x - y, but that is the same as 360-x-y-60 = 300-x-y).

[@ivanabcdefg9375]

This solution is more satisfying than the one in the video because it is symmetrical, and because after the mirroring the remaining steps are intuitive. One note, in step 2 the triangle sides are slightly smaller than 2x,2y,2z after the flattening; they are actually sqrt(3)*x,sqrt(3)*y,sqrt(3)*z due to the projection down from the 30 degree angling. But still, it's a similar triangle, and so the angles are the same as if its lengths were x,y,z. QED.

[@MintAvhnDaoeAdjd]

I can't figure out what you mean by "flattening away": do you mean by extending alternate sides of the hexagon? If so, I don't think it seems intuitive that the resulting triangle will be similar to the one you're looking for. Reading your step 2 a few more times, it sounds like you might be truncating the hexagon into a triangle circumscribed by it, but again it doesn't seem obvious that that triangle is similar, or even that you have to cut pi/6 or pi/3 off each angle to do it.

[@1234s6]

You can check and confirm that this angle formula is true; Let M be the midpoint of the equilateral triangle. Connect each vertex to M with a line. Since its the midpoint, the angles; x, y, and (360-x-y) are all 120°. Now since each line has the same length, when removed from the equilateral triangle they form a new equilateral triangle. So all the angles are 60°. So: x-60 = 120-60 = 60 y-60 = 120-60 = 60 (360-x-y)-60 = 300-x-y = 300-120-120 = 60

[@bradenf]

This video is not color-blind friendly. This is why I had trouble in Electronic Engineering. I'd much prefer labeling line segments A, B and C instead of color coding them.

[@futuredystopia5935]

Differently dashed lines may be easier to parse.

[@bradenf]

Future Dystopia Agreed! I don't care if it's combined with color coding as long as it can be made intelligable to the rest of us.

[@VolantData]

Totally, I'm colorblind and I got kind of confused

[@just-dl]

when doing graphics of this variety, I always did line-type differences with colors....because they often got photocopied. Back in the good old days, your photo color options were "black and white." Not even gray scale. So, my premise was always, draw it so it can photo copy well. A concept that I stick with today, but, I find it less necessary. Braden, thanks for the reminder that visual discimination aids are always a good idea....even if color copiers and printers are readily available.....

[@davem3953]

I've read that 6 percent of the population is color blind. That's a lot of people.

[@amrutesh1000]

There is another solution. Whatever is the formula for the triangles angles, has to be true no matter where the point is. Thus it has to be true for centroid - x & y both are 120 and angles in triangle are 60. A bit of trigonometry for a isoceles right triangle inside - you will find y = 105 and x = 150 for eventual angkes to be 45 45 90. So if x = 150, y = 105, a=b=45 If x=120, y=120, a=b=c=60. So you can work towards the solution - however the suggested solution is far more elegant

[@ronginzler6600]

I figured it out by cutting up the original diagram into three triangles and putting two of them back to back with two of the sides of the original triangle coincident. But only after I'd spent several hours on it and woke up in the middle of the night with the answer.

[@fengshengqin6993]

There is another way to solve this problem. And this is more elegant and more easily understandable. You can flip the three little triangle to the outside of this equalateral triangle, then you can get a hexagon ，which has the edge length as the same as the original values. Three vertexes of this hexagon is 120 degree. You can link the other three vertexes and get a new triangle . The edge of this new triangle is squaroot3 times of the original edge length values. So this new triangle can be as the requested one. Very easy to see that ,the angle of this new triangle is x-60 ,y-60 . Problems solved . Like I said ,more elegant and more easily to understand !

[@xyz.ijk.]

Colors presented a real problem for me, and you don't ever seem to be able to take that into account, but other than that your explanation was masterful.

[@druariel]

Do I get a special prize for solving it in a different way that’s ten times harder and longer 😂

[@garethb1961]

Me too!

[@giantbonsai]

I think, You actually can brute force this. The two angles allow you to get all angles of the three triangles you have devided the equilateral triangle into. You can then calculate the lengths of the three sides the final triangle is composed of with the sidelength of the equilateral triangle as a variable. Take the longest and put it along an arbitrary axis. Now define two circles centered at the end points with the length of the remaining sides as their radii. The point of intersection will be definded by two angles along the circles which can be obtained by solving for this point. These are two of the angles. Now retrace your calculations to express these in terms of the original angles. It takes some trigonometry but it should be doable.

[@wes9627]

If you have computed lengths of the three sides then you may use the Law of Cosines to compute the three angles.

[@BigBossHuntelaar]

Beautiful! Such an elegant solution

[@bonapartebrothers4232]

This is what i call elegant.Good sum with an elegant construction.

[@alexandermikhailov2481]

A very elegant solution, bravo!

[@gedlangosz1127]

First I tried a special case of the "random point" lying at the midpoint of one side. By constructing the new triangle from this special case, and analysing the angles, I guessed what the solution was. To prove it was then straightforwards. I constructed a new equilateral triangle in the original diagram, using one of the lines touching the random point as its base. Now join the naked vertex on the new equilateral triangle to the nearest vertex on the original triangle. It is clear that we now have constructed a triangle congruent to one in the original diagram. It's then easy to deduce the correct result. BTW call the original angles x, y & z - it makes everything a bit more elegant.

[@darrenl3289]

You can also take a hint from the original stated direction: choose a random point inside the equilateral triangle. Knowing you can choose any point implies the solution will work from any point so you can choose the easiest starting point, the center. This splits the original triangle into six 30 degree angles at the outer vertices and three 120 degree angles around the center point. Now when you rotate the original triangle, you have practically all the angles already calculated. This allows you to find the ratio to calculate the final angles, which can also solve for any other random point. Remember, if given the option, work smarter, not harder.

[@NoNameAtAll2]

"Did you figure it out?" MYD: "Nope"

[@arikwolf3777]

I didn't even try.

[@lingarajnayak1771]

Keep this problem a try

[@jyotirmoy-paul]

Literally Ingenious Solution! 😯

[@sunriseinspector]

I worked it out by "brute force". I called the third angle in original triangle, where the three colored angles meet, z. I called the angles in the triangle that was formed by the three colored sides a, b, and c. I was able to show that for my choice of a, b, and c: y-x=b-a z-y=c-b z-x=c-a. The proof is messy and long, I'm not typing it in, but it was just high school geometry and simple algebra. I then used these three equations along with x+y+z=360 and a+b+c=180 to solve for a, b, c.

[@lelouchvibritannia1410]

Bravo!! Now it's time to solve for x and y

[@alcodark]

I want to see the "brute force" solution though...trigonometry and all...

[@luke46219]

Here's my brute force solution, using the picture with angle x between blue and purple line segments and angle y between blue and yellow line segments. Assume angles are in degrees. First some definitions. Define point C as the point in the middle with angles x and y next to it, point D the left vertex, point E the right vertex, and point F the top vertex of the equilateral triangle. Then let xD be the angle FDC, that is, the angle between the black line segment with endpoints being vertices F and D and the blue line segment with endpoints being vertices D and C. Also let z be angle FCE and zE be angle FEC. Finally call the length of the purple line segment connecting points C and F length P, the length of the yellow line segment connecting points C and E length Y and the length of the blue line segment connecting points C and D length B. Also, let each side of the equilateral triangle have length 1, since the actual dimension would only be important for finding the lengths of the colored line segments. The law of sines gives sin(x)/1 = sin(xD)/P. Also, sin(z)/1 = sin(zE)/P. So sin(xD)/sin(x) = sin(zE)/sin(z) since each side of this equation equals P. It's easy to tell sin(z) = sin(360 - x - y), but it takes some calculation to find sin(zE) = sin(y - 60 - xD). Now let S = sin(x)/sin(360 - x - y) so the next few equations will be simpler. With this, you can write sin(xD) = S*sin(y - 60 - xD). Use the trigonometric property sin(a-b) = sin(a)*cos(b) - cos(a)*sin(b) to give sin(xD) = S*(sin(y-60)*cos(xD) - cos(y-60)*sin(xD)). Rearrange to give sin(xD)*(1+S*cos(y-60)) = cos(xD)*S*sin(y-60). Then tan(xD) = S*sin(y-60)/(1 + S*cos(y-60)), or xD = atan(S*sin(y-60)/(1 + S*cos(y-60))). Everything in this equation is in terms of the given angles x and y so you can solve for xD. Make sure the calculations are for degrees. Now you can easily write equations for the lengths of all three colored line segments, assuming the equilateral triangle has side length of 1. P = sin(xD)/sin(x). Y = sin(60 - xD)/sin(y). B = sin(180 - x - xD)/sin(x). After all this, you can use the law of cosines to solve for the angles of the triangle made by the colored line segments. The law of cosines is just a general form of the Pythagorean theorem, a^2 = b^2 + c^2 - 2*b*c*cos(A) where a, b, and c are the lengths of the sides of the triangle and angle A is the angle opposite side a. Then A = acos((b^2 + c^2 - a^2)/(2*b*c)). Hope this helps.

[@cyberpilot6512]

study the law of cosines - dont try it without a good cup of coffee.

[@nickwilson3499]

Freedom Fighter when do you learn law of cosines? Is it in algebra 2? I just finished geometry and I’ve never learned it.

[@Tehguil]

Very interesting problem! I'm proud to have achieved to solve it :)

[@damontan4749]

To people that have trouble understanding the question. Seriously? Even after watching the whole video, listening to his explanation in addition to the DIAGRAMS drawn? Great job on being smart.

[@mike1024.]

I thought about making a copy of the bottom triangle and lining up the congruent sides, so that seems to be my head going in that direction. I didn't work out the details before watching, but now I wish I would have tried!

[@Adityarm.08]

I did it with a completely different method. Tough to put in words. But I considered symmetrical cases, moved the point to the top vertex then splitted the side of concern on top and slid it down till it touched the other side of concern. Then joined the two halfs! And there you have it 🙂

[@bh8642]

The problems were NOT used to "discriminate against groups fo people" - they were used specifically to discriminate against Jews in the USSR who wanted to go to university there. Some months ago this same page said so explicitly, but apparently trolls got the upper hand.

[@sharonminsuk]

Thank you, bh. I wasn't aware of this bit of history. It's good to get the real story.

[@alexeytsybyshev9459]

After a couple of minutes, I came up with a different solution. While trying to find the triangle with these segments as its sides in the picture, I took mirror reflections of the point with respect to the sides of the original triangle. Now if you look at the line segments connecting them, they are the'the bases of three similar isosceles triangles with 120 degree angles opposing them, and our original line segments as their sides. Therefore these 3 reflections form a triangle that is similar to the triangle formed by the 3 line segments we are interested in, and it remains to calculate its angles, which is easy.

[@yolololfun5341]

Did the same thing

[@akaRicoSanchez]

I think you mean 'the bases of three similar ISOSCELES triangles with 120 degree angles opposing them' :) But nice solution!

[@alexeytsybyshev9459]

Thank you, I mixed them up. I fixed it now.

[@FundacjaCrown]

I have found similar solution to yours

[@geetavagvala4473]

Pesh, there is a diff solution to this problem. But I could on,y come up with this solution only because I now know the actual answer. Anyway let me give a new variation of the problem in which there is also this second solution hidden in it. If you take any scalene triangle triangle say sides a , b, c.. Draw opposite side to parallelogram from say one vertex(ac).. I.e. draw a parallel line of size equal & parallel to b?Then draw an equilateral triangle with sides b from that vertex away from the original triangle with one side being this new parallel line to side b. Like wise from same vertex draw another equilateral triangle with each side as c away from the original triangle with one side being c from the original triangle. Can you prove a triangle drawn from vertex (ab) from the original triangle and new vertices of two equilateral triangles is also an equilateral triangle (clue vectors)... Now this is the second solution to the original problem..

[@geetavagvala4473]

Btw, this is Venkat Vagvala didn’t realize my iPad is logged in my wife’s name.. Anyway if you are interested I. this second solution, pls drop me a note to [e-mail]. I can send you some penciled drawing to illustrate the second alternative..

[@jorgensenmj]

I solved it easily because I already knew the "equilateral" triangle trick from other geometry problems. Whenever I see the problem involves "equilateral triangles" the solution almost always involves a rotation of 60 degrees. I would be careful giving people credit for an "ingenius solution" when the solution has probably been around since the time of the pyramids.

[@emekdulgeroglu3914]

I have found a harder way to solve the problem, which is using a=b*c*sin(x) formula. If you use it for each three side, any angle could be calculated in terms of sin(x), sin(y) and sin(x+y). However the method in this video is much easier than the method I used. Great work!..

[@MrQwefty]

Damn.... that totally got me!

[@hippophile]

Very neat solutioon. As a statement of the solution, I prefer to say there are three angles in the interior of the first triangle: x, y and z. Then the three vertex angles of the new triangle are (nice and symetrically) x-60, y-60 and z-60. It just seems more elegant to me...

[@jthegenius]

​Here is a solution using the properties of complex numbers. Consider the diagram shown, e.g., at time 1:00 min​. Imagine the whole construction being on the complex plane. Let the random point inside the triangle (from where the colored lines emanate) be the origin O on the complex plane; nothing prevents us from making this choice. Now label the three vertices of the triangle as A, B and C such that the blue line is OC, the purple line is OB and the yellow line is OA (with appropriate directions implied in each case). Thus according to the diagram, the magnitude of angle BOC = x and magnitude of angle AOC = y. Let us denote the complex numbers OA, OB, OC and BC by a, b, c and z, respectively. Since ​the triangle ABC is equilateral, the complex numbers CA and AB are then z*w and z*w^2 respectively, where w = exp (2 i pi/3) .... (Eqn.1), is ​one of the cube roots of unity. Clearly by the above choice, the lengths of AB, BC and CA are all equal to |z|. The fact that the three complex numbers add up to zero (and hence forms three sides of a triangle -- the​ "triangle rule of addition of complex numbers" in action!) follows readily from the well-known identity 1 + w + w^2 = 0 .... (Eqn.2). We will next apply the "triangle rule of addition of complex numbers" to ​the sides of each ​of the three smaller ​ triangles, namely ​triangle BOC, ​triangle COA and triangle AOB, separately. This yields, b = c - z .... (Eqn.3), a = c + z*w .... (Eqn.4), a - b = -z*w^2 .... (Eqn.5). (Eqn.5) can be obtained by subtracting (Eqn.3) from (Eqn.4), and is therefore redundant.​ However, ​a different linear combination of (Eqn.3) and (Eqn.4) eliminating z yields (-w^2)b + (-w)a = c .... (Eqn.6). Let me point out that the right hand side of (Eqn.6) has been simplifed with the help of the identity in (Eqn.2). Now imagine rotating the complex numbers OA and OB around the origin O. We do so by keeping the origin O as well as the complex number OC fixed. Let us keep rotating both OA and OB in this manner until OA and OB form two sides of a parallelogram with OC as the diagonal in between them. Denote the new vertices, i.e. the new positions of the tips of OA and OB, by A' and B' respectively. The fact that OA' and OB' form the sides of a parallelogram with OC as the diagonal in between implies that the triangles A'OC and COB' are two identical copies of the triangle whose angles are being sought for. For convenience, denote the new complex numbers OA' and OB' by a' and b', respectively. Since OA' comes from OA under a rigid rotation around the origin (i.e., magnitude stays unaffected), by an angle Y in the clockwise direction (say), we may write a' = a exp (-i Y) .... (Eqn.7). Likewise for OB --> OB', we may write b' = b exp (i X) .... (Eqn.8), where X is the magnitude of the angle through which OB is rotated counter-clockwise (say) to turn it into OB'. Thus, the magnitude of angle B'OC = (x - X) and magnitude of angle A'OC = (y - Y), by the above considerations. Note, however, that the angles B'OC and A'OC are two of the three angles to be found (the third one then follows from the fact that the sum of the angles of a triangle is 2pi). Hence, our goal is to find the angles X and Y. Now, by the "parallelogram law of addition of complex numbers", we have b' + a' = c .... (Eqn.9), as the condition for the formation of triangle whose angles we are after. By (Eqn.7) and (Eqn.8) this is equivalent to b exp (i X) + a exp (-i Y) = 1 .... (Eqn.10). Let us now compare (Eqn.6) with (Eqn.10). If these two equaitions were linearly independent, we could have solved for a and b uniquely from them in principle. But this is impossible since the point O is a random point, and there are infinitely many possible ways to choose the point O inside the triangle for which our analysis holds. Consequently there are infinite choices for a and b for all of which the above considerations apply. Hence, (Eqn.6) and (Eqn.10) must be the same equations. Therefore in particular, the coefficients of a and b must agree in these equations. Comparing the coefficients, we then have exp (i X) = -w^2, i.e., X = pi/3, exp (-i Y) = -w, i.e., Y = pi/3, with the help of (Eqn.1). This agrees with the solution presented in the video, namely, two of the three angles are pi/3 (= 60 degrees) less than the angles x and y as defined in the original figure.

[@miikey_lol]

Hey, love these types of videos!

[@alecdybiec2535]

If you attempt the problem with a pythagorean theorem approach to get a side length and a backwards pythagorean theorem to get another side length, while not knowing the lower portion of the hypotenuse of 10 to begin with, you can deduce that the figures given are impossible for a right triangle simply because all the logical values you could assign to that lower portion to begin the pythagorean theorem strategy would result in one of the legs being longer than the hypotenuse, which we all should know can't happen for a right triangle.

[@jimlocke9320]

There is no need to rotate the original triangle. Just construct the blue equilateral triangle on its own and construct a line segment from its vertex outside the original triangle to the top vertex of that triangle. Then, we don't need to prove that the blue triangle is equilateral. Now, let's label the vertices for reference. The original triangle vertices are: lower left = A, upper = B, lower right = C. The random point is D. The constructed equilateral triangle is ADE. Note that angle EAB = angle DAC because both are equal to 60 - angle DAB. Triangle EBA is congruent to DCA by side - angle - side. Proceed with Presh's solution for the angles of triangle DBE (about 5:45 into the video). Another observation: Let's label the original third angle at the random point, angle BDC in my drawing, as angle z. Then, x + y + z = 360. Presh's "final angle", which I call EBD, becomes z - 60. So, the 3 angles in the triangle formed by the blue, yellow and purple line segments are each 60 degrees smaller than the angles at the random point formed by the same colored pairs.

[@ericaroozendaal629]

Yes I did it! I enjoyed labeling the third corner z in order to not having to write 360-y-x but z-60

[@xexpo]

I got it straight away. Incredibly easy if you visualise the lines coming together and you notice how the new angles related to the original.

[@dailyfunny2399]

Here is my thought flow, hopefully it is helpful to someone: Let z be the third angle so that x + y + z = 2 pi. (1) First consider the case when x = y. In this case we will get an isosceles triangle, and the top angle w would be some function of z, say w = f(z). (2) Experiment and see that when z = pi/3 (a limit case), w = 0; when z = 2pi/3 (the case when the random dot is actually the center of the original triangle), w = pi/3; when z = pi (another limit case), w = 2pi/3. The last case takes a bit of computation. We actually get an isosceles triangle with sides 1, 1, sqrt[3], then it is not hard to find out the top angle w = 2pi/3, say by using the height to divide the isosceles triangle into two. (3) Now observe that f(z) is very likely to be z - pi/3 by the three values f(pi/3) = 0, f(2pi/3) = pi/3 and f(pi) = 2pi/3. (4) w = f(z) = z - pi/3 leads to the thought that if one rotates the yellow (or blue, or purple) line by pi/3, or 60 degrees, one might construct a triangle like the one shown in the video.

[@andvil01]

Put it up in Geogebra. Put a random point in the triangle. The distance to the point a b c. Made the new triangle of the sides a b c with circles with radius a b and c. Set the angles around my point and in my new triangle. As I moved the point around I saw the angles was always 60 degrees less. There is a logic too it as well. The sum of the angles around the point is 360. And in the triangle 180 degrees. As it doesn't matter where the point is, those 180 degrees should be taken even off the three original angles. So -60 degrees on each. It's not a proof, but it is ways to get to the answer.

[@faresamir4436]

Hi. i found the solution astonishing but i have another solution in mind that doesn't involve the trick of rotating the triangle. i am not sure about it so would you please tell me if it's true or false? Can't we say the angle between the yellow and the purple segments (let's name it z for example) is 360-x-y? then we would be able to use Al Kashi's law of cosines on the angles x,y and z in each triangle. since the big triangle is equilateral we can link the three equations we got because the sides of the big triangle are equal. after that we would have a system of three variables a b c which are the lengths of the colored segments with the three equations. then we'll be able to solve for the three lengths according to the cosines of the angles (that according to the problem are known) then, knowing the lengths a b and c, we can apply al kashi's law of cosines again on the triangle we want to make from the three colored segments. we will obtain the cosines each angle of that new triangle and thus we use cos^-1 to obtain them. i didn't obtain the solution of this method yet to check because i m not familiar with systems of three equations yet. i hope u look into this method and tell me if it is right or not. thanks!

[@Pengochan]

I figured it out. What i did first was to think about possible relationships between x,y,z=360°-x-y and the angles in the resulting triangle. x,y,z add up to 360°, the angles in the triangle add up to 180°, so the angles could be halved or it could be -60°. It is also clear, that the relationships between x and x', y and y' and z and z' must be all the same, where the v' is the angle between the same colored angles in the triangle. Of course there could be a more complex relationship, but as the solution is supposedly elegant (and the video not that long) let's look at simple things first. Immediately the x -> x-60° variant seemed more appealing, since there are 60° angles in an equilateral triangle. I also know these rotation completion tricks from other similar proofs. Those often work by parallel shifting sides, but a pure translation wouldn't do here. I considered the extreme case where the point is very close to a corner and e.g. the yellow side very short. then x would be a little larger than 60°, but the angle between the blue and pink side of the resulting triangle very small. I then considered the rotation as in the video. This results in a z-60° angle in the yellow pink corner, but i have a quadrilateral with two blue sides. It took me a second look to see that the two blue sides enclose a 60° angle, so they can be completed to an equilateral. Tl;dr.: "Elegant" short solutions are often kind of simple, so look for simple stuff first. Look for "symmetries" (here none of the angles was "special"). Consider extreme cases. In geometric puzzles move stuff around.

[@antonfelix1]

I solved it another way using trigonometry: I started to solve for all the angles in the 3 triangles created by the lines going to the random point. In these triangles 1 angle is known for each: x, y and 360-x-y. to solve for the rest of the angles in these 3 triangles: let one of the unknown angles in each triangle represent a,b and c. that way we get the three following equations: 180 = a + (60 - b) + x 180 = b + (60 - c) + y 180 = c + (60 - a) + (360 - x - y) now solve for a, b and c and get all the angles in these three triangles. next, assume the common side in these triangles is of length 1 (the side that makes the equelateral triangle) and solve for the length of the lines going to the random point (only two of these triangles are needed for this). using the law of sines solve for the angles asked in the question from the known side length

[@prathampaul7897]

Truly an Ingenious sol.

[@IanNewYashaTheFinalAct]

When I saw the thumbnail for this video, I thought the triangle was supposed to be a tetrahedron a.k.a. A four-sided die

[@guymiller5064]

your youtube channel is incredible by the way

[@fluffyorange647]

What’s even cooler is that if the longest side is the sum of the squares of the two sides, the angle between the two shorter sides is always 150 degrees!

[@SachinthaSilva]

What if we take out the internal three line segments as 3 outward directed vectors (at the same scale) and form the corresponding diagram. Since the initial lines (vectors) intersect at a static point, the newly formed diagram by these lines (vectors) must be aligned with a complete polygon (due to the equilibrium of three inclined vectors). That proves the internal angles of the vector polygon (in this case, a triangle) are (π-x), (π-y) and ([x+y]-π). As similar to the "Lamé's theorem".

[@susmitamishra8436]

By induction: We take some special cases WLOG(without loss of generality) let point be on midpoint(case1) here the lines from vertices will be a/2,a/2,√3a/2(let side of triangle be a), acc. to question here

[@handle535]

This is how I did it, although it feels less rigorous than the answer in the video.

[@trueriver1950]

at 5:58 you have already solved it - there are two ways to see this. Firstly the point is arbitrary so you simply move the x and y round the point. Alternatively rotate the original triangle by 60deg about EACH vertex.

[@coinagevolume]

i feel its a mistake @ 5.31 min cos the angle which u considered 60 degree saying it becomes equilateral triangle , it should not include angle x in it , it means it itself is 60 degree and x is different from the trianlge formed , correct me if i am wrong

[@angrytedtalks]

I thought it was obvious. Turning it round just makes me dizzy. I am colour blind - could you label the triangles in question?

[@ernestomamedaliev4253]

I would use the sine theorem for searching the distances, and the cosine theorem for finding the angles of the new triangle; but this solution is much more elegant!

[@luigipirandello5919]

Very beautiful solution.

[@Vcimdarf]

Wow! I did not find this very neat final formula, but I managed to solve it effectively with trigonometry. I ended with a gigantic formula at the end, which probably simplifies to the right answer (it works, so it has to). I named the three internal lengths a (purple), b (blue), and c (yellow), then I introduced two new angles theta and phi, at the extremities of the purple and yellow segments, inside the x and y triangles. On the one hand, we have x + y + theta + phi = 300. (1) On the other hand, we have (from the law of sines), sin(theta) / sin(x) = sin(phi) / sin(y) (2) You can rearrange (1) a bit and apply sin at both ends to get sin(theta) = sin (300-x-y - phi) = -sin(phi)*cos(300-x-y) + cos(phi)*sin(300-x-y) Divide each side by sin(phi), you get an expression with cot(phi) only sin(theta) / sin(phi) = -cos(300-x-y) + cot(phi)*sin(300-x-y) (3) Apply (2) into (3) to get phi from x and y only, then into (1) to get theta from x and y only. Once you have theta and phi, all the quantites a, b and c can be found with the law of sines. Then, the angles of the final triangle can be derived from the law of cosines. The final formula is very long but you can plug x and y into it and get a straightforward answer.

[@VolodymyrRushchak-k6l]

It's a REAL beautiful problem!

[@bibhuprasadmahananda6986]

Mind blowing problem, this is one of those very few problems that I wasn't able to figure out.

[@dd-explores]

Truly an elegant solution

[@paintingjo6842]

Wow, I'd been so close to figuring it out, I thought of rotating the parts around a vertex, but unfortunately overlooked connecting the point and its duplicate. Ah, the little things you don't notice when working on paper.

[@Skandalos]

I used geogebra instead of paper and found it :)

[@swapnil3990]

It's easy Let the sum of the three angles, X , Y and the known angles A be 360° in the first case. Now, for the second case, the sum of internal angles should be 180°. The sides are unchanged, therefore the ratio between the angles and hence the sides will be constant. Thus, the final angles will be X - 60°, Y - 60° and A - 60°. (This is because there are three angles and each angle should be reduced equally by 60° to reduce the angle of 360 to 180°.

[@Shellova]

Oh i actually figured this one out :D nice one!

[@danradu2833]

An equivalent problem ... and implicit solution: Let the equilateral triangle be ABC and the internal point M. Let P, Q and R be the perpendicular projections of M on ABC's sides. Show that triangle PQR is similar to the triangle formed with MA, MB and MC and that the similarity ratiois sqrt(3)/2. For an arbitrary triangle ABC the angles of the resulting triangle PQR can obtained by subtracting the the angle of the vertex from the angle at M opposite to the vertex(e.g. x - A, y - B and z - C). What's more, it can be easily generalized for any regular n-polygon ... the projections of a random point M on the polygon's sides form a polygon whose sides are in a constant ratio (sin pi/n) to the segments determined by M and the regular polygon's vertices.

[@dneary]

A related problem, with a similar solution, is to find the point P inside any triangle ABC that minimizes |AP|+|BP|+|CP|

[@wes9627]

I haven't looked at the solution yet but I was thinking along the lines of using the Law of Cosines. Assume that the sides of the equilateral triangle are unity. Since angles x and y are given, the third angle z = 2*pi - x - y is also known. Let a, b, and c represent lengths of the three lines from the point to the vertices. Then three quadratic equations using the Law of Cosines may be expressed in terms of cosines of the three known angles x, y, and z and the unknown edge lengths a, b, and c. If there is an elegant way to directly solve these quadratic equations for a, b, and c then the Law of Cosines may again be used to evaluate the three angles of the new triangle in terms of edge lengths a, b, and c. If I can't find an elegant way to solve these three quadratic equations for edge lengths a, b, and c, I would use an iterative method like Newton-Raphson iteration to compute them.

[@AaronHollander314]

I actually got it a different way. First, I assumed the random point was infinitely close to one of the vertices. Therefore, the angles would be ~150°, ~150° & ~60°. And the newly formed triangle would be ~90°, ~90° & ~0°. (~ meaning very close) Then I moved the random point to the center of the original triangle with all three resulting angles being 120° and the newly formed angles being 60° (using those segments would yield a similar triangle to the parent). So, wherever I started the middle angles (x, y & z) would be 60° greater than the angles of the segment formed angles (x', y' & z'). x' = x - 60° y' = y - 60° Solving for the third angle the same way as in the video z' = 300° - x - y I do not think this is a proof since I just guessed that every point in between (and all around) would follow suit. But it yielded the right answer (I think).

[@souravmishra5828]

completely stumped!! tried producing the line segments to the random point but got nowhere. thanks.

[@adityakossambe5265]

I just plotted a typical situation in autocad and the answer was rather obvious. Although its not the right method to solve this and the method displayed by Presh is rather ingenious. To put it like a typical solution considering the three angles formed at any given point inside an equilateral triangle be x , y , z then the angles formed by the new triangle will always be (x-60) , (y-60) , (z-60)