A very hard admissions question in India

[@MindYourDecisions]

This problem is adapted from an examination paper from the Chennai Mathematical Institute (CMI), a premier research and education institute in India. It was sad to read that Professor C. S. Seshadri, the founder of CMI, passed away this week at age 88. In algebraic topology, the Seshadri constant is named after him. K. VijayRaghavan, Principal Scientific Adviser to the Government of India, writes: "C. S. Seshadri's lasting contribution is that he has ensured there will be many more like him from the CMI, and from all over India. A life in mathematics, music, institution building, and humanism. Worth understanding and its core values worth emulating, no matter what we do." twitter.com/PrinSciAdvGoI/status/1284406586140528640 C.S. Seshadri on Wikipedia en.wikipedia.org/wiki/C._S._Seshadri Seshadri constant en.wikipedia.org/wiki/Seshadri_constant Bhavana magazine. From Proofs to Transcendence, via Theorems and Rāgas. C.S. Seshadri in Conversation bhavana.org.in/proofs-transcendence-cs-seshadri/ C.S. Seshadri passes away at 88 theprint.in/theprint-profile/cs-seshadri-padma-bhushan-and-chennai-mathematical-institute-founder-passes-away-at-88/463840/

[@worobiul8627]

English is not my first language so i did not understood that much. But it is a good puzzel. I will give this to my math teacher.

[@shrivatsboi]

May his soul rest in peace

[@ybhan1]

I used another method, extend BF and EF to cross CD at G and H, there will be two pairs of congruent triangles EDG-EFG and BCH-BFH, in the right triangle FGH, we can calculate the hight (2) by using the Gougu theorem (勾股定理)

[@ybhan1]

I think this method is beautiful in balence

[@lightlysal]

I really think you should do away with calling it the Gougu theorem, because - "Pythagorean Theorem" is more widely understood and is just more practical - The origin of the theorem is debatable - It's not unethical to refer to something differently than the original (which is debatable anyway), there's a reason languages have different interpretations of other words (otherwise we'd be calling Mung Beans Green Beans), it's not offensive - Chinese students call it the Pythagorean Theorem in English, because it's the most widespread version Of course you can feel free to call it whatever you like, but if you consider it either better or "correct," I'd have to object.

[@Beebo]

I have a different solution: Use equations of a circle by making the point D (0,0) 1. x^2 + (y-5)^2 = 25 2. (x-10)^2 + (y-10)^2 = 100 Make them equal to each other and find the coordinates of intersection. You know one of them is (0,10) and you solve to find the other one. Which is (4,2). y = 2 so F to DC is 2.

[@thewaytruthandlife]

I agree, thats the method I used as well... congrats I admit I used excel to find the solution but OK ...

[@antonijesubotic3819]

It is nice aproach, but still it is too easy problem

[@gokuvegeta9500]

@@thewaytruthandlife But how do we know that the intersection of both circles is at point F ???

[@Beebo]

@@gokuvegeta9500 EF has a length of 5. And the center of the circle is at E. The radius of the circle is also 5. BF has a length of 10. And the center of that circle is B.

[@gokuvegeta9500]

@@Beebo Ahhhh, now this was the explanation I was looking for! Damn it was so obvious, but I somehow missed it 🤦🏻‍♂️🤷🏻‍♂️

[@spandanhalder9967]

Presh intentionally does the problems with geometry 1st even if there are other shorter methods, so that eventually he can deliberately use and say GOUGU THEOREM.

[@nabelalifxiiak4565]

Yes lol, he has been waiting for this moment

[@Om-id1qr]

XD

[@spandanhalder9967]

@@nabelalifxiiak4565 Exactly😂

[@CheyenneAnastacia]

HAHAHAHAHAAHAHAAHHAA

[@shravberri]

Yes, apparently he LOVES to call it something chinese(don't take it the wrong way, but we kinda have a love-hate relationship with China), when he himself is Indian, who according to me should be calling it the Baudhayana theorem (cuz indian roots duh, but idek anymore cuz he doesn't seem to acknowledge it). Pure disgrace

[@unknownwriter1507]

I found a much more straightforward and simpler soln for this:- Draw a line LM passing through F parallel to DC. Taking distance of F from DC as x and using pythagoras theorem we get- √(5²-(5-x)²)+√(10²-(10-x)²)=10 =>x=2,10 and since obviously the distance of F from DC cannot be 10 as then it would be equal to the side of the square, it would be 2

[@jiabhargava]

OMG! I used this method too and it's so much easier. Solved the square roots by squaring both sides twice. I was wondering why this method wasn't even mentioned in the video

[@techsandbulls1869]

YES THE SAME THING I DID AND THOUGHT I MIGHT HAVE GOT IT BY FLUKE BUT ITS RIGHT AND ACCURATE

[@CircuitShorts]

Same here

[@lithograhp]

FEL is similar to BFM and EF/FB=10/5=1/2. It is very easy to find EL=3.

[@ranjanakushwaha2534]

I also used the exact same way 🎉

[@donaldasayers]

This is a solution to the much harder problem: only by folding, fold a square of paper into five equal areas.

[@aryannagariya6027]

Simple fold it into strips

[@donaldasayers]

@@aryannagariya6027 How do you know where to make the first fold?

[@aryannagariya6027]

@@donaldasayers from edge

[@donaldasayers]

@@aryannagariya6027 You need the answer to this problem to show where to make the first fold. Obviously you fold it into equal strips, that's trivial. What are you trying to say?

[@aryannagariya6027]

@@donaldasayers just fold it into strips in the intuitive way, take a edge and fold 1/5 length of side, do it over and over and you will end up with 4 line of square which divides it into 5 parts

[@mayankd8362]

Use trigonometry answer in 1 minute Find angle FBC using cot(2theta) Where tan (theta) =1/2 Then find 10-10cos(2theta)

[@rogernewcomb9014]

You meant sine I believe (you're subtracting the opposite side from the whole square side, not the adjacent side) otherwise, pretty much exactly what I did.

[@andrewadoranti1423]

Yes same here, every solution is so overcomplicated for this question

[@nisiu007]

I've watched hundreds of your videos and this is the first one that I managed to solve by myself. I used trigonometry, but in a slightly different way and I got 1.989 but that's because of approximations

[@patelraj1029]

Me too

[@wallstreetoneil]

i got 1.999 (i used more decimals)

[@geoninja8971]

@@wallstreetoneil Guys, keep the values in your calculator, don't round at all, and get 2.0000000 :)

[@masterplayer5982]

I used a cheap calculator so I doubt that it would ever get 2 😅

[@mbart2066]

Same here

[@burlakoff_av]

So, there's another meme: Nobody: Absolutely nobody: Presh Talwalkar: Now we can use the Gougu theorem for this triangle. Really? But why Gougu?

[@JohnDlugosz]

Why indeed? Euclid gives an axiomatic proof 200 years before Gougu. My wife is Chinese, and doesn't know the English names for things she learned in gradeschool. Sharing a puzzle with her, she tried to describe the method and I said "Gougu?" Yes! and continued with the puzzle. More people speak Chinese as a first language, so I think the term 勾股定理 is actually the more common.

[@spiderjerusalem4009]

because he's so obsessed with math, even the history lessons of math

[@virajagr]

@@JohnDlugosz I guess there are more English speakers than Chinese.

[@haggisllama2630]

@@virajagr if you look at numbers exactly, not taking into account how widespread the languages are. Chinese/mandarin is spoken by a lot more people. However, most of that is in China and the countries around it. English is far more widespread in where it is spoken.

[@mohammedjawahri5726]

@@JohnDlugosz not just that, but to follow up, theorems are very commonly not named after whoever proved them, names are just that, names. We call theorems what they are because they became associated with a figure, some theorems are known to have been named after "investors" who literally paid to have the theorems named after them (e.g bernouli's iirc?) Some other theorems are named by whoever popularised them, etc. Our professor once gave us a funny segment where the premise basically was "no matter whose name is on the theorem, chances are some arab, chinese,indian or greek guy discovered it some couple of centuries ago" lol

[@WillRennar]

I found the angle of AEB, doubled it (since BEF is the same angle), and subtracted that from 180 to get angle DEF, which was 36.87, the angle in a 3-4-5 right triangle between the opposite and the hypotenuse. Drawing a 90º angle from DE to F (point H on your video) gives us a 3-4-5 right triangle HEF. Thus, since EF is 5, that means that EH is 3 and HD, which equals F's height from the base, is 2.

[@kostasl1808]

I did it with conic sections and x-y axis system. I used the circle formula and took 2 circles: C1: center (0,5) and r1=5 & C2: center (10,10) and r2=10 Thus we get 2 formulas: x^2 + (y-5)^2 = 25 (x-10)^2 + (y-10)^2 = 100 by solving the system for 0

[@jreddy5234]

There is another way to solve it using basic vectors concept , here we know 2θ = 2tan_inverse(1/2) so by simplifying it we get 2θ = sin_inverse(4/5) which equals 53°. So the projection of FB on BC would be “10sin53°” which equals 8 hence FC is “2”

[@Om-id1qr]

It's easy but seems confusing at first. Thanks Presh for bringing such mind refreshing problems!

[@shiwenhuang6997]

I have a different approach. If I fold the other end "C" to "F", I can obtain a right triangle EDP (P is a point on side DC when this folding occurs). And let me assume PC = x. And it is obvious PC = FP = x. I can simply obtain the relationship between x and square's side through ED² + DP² = EP² because EP = EF + FP = 5 + x. So I will get x = 10/3 = FP. Then if you look into triangle EDP, you will see another similar triangle FQP (Q is a point on DP when you draw a line from F in parallel to ED and that's the distance from F to DC). Because they are similar triangles, we have FQ/ED = FP/EP => FQ = (5*10/3)/(10/3 + 5) = 2. Too bad KZbin cannot allow to post an image, otherwise it would be much easier to show you my solution graphically.

[@bachpham6862]

Draw the line DF and FC, then you will split the square into 5 triangles. You can calculate the area of triangle DEF and BFC as you know the length of 2 of their sides and the angle in the middle. Subtract the are of those 2 and AEB and EBF from the square, you will be left with the area of DFC, which you can calculate the height since you also have the base.

[@peterromero284]

This video is in a foreign language that’s exactly identical to English in every regard except the translation of Pythagorean is Gougu.

[@ezequielcallegari2644]

Why are you so obsessed with the name of the theorem? He calls it Gougu because he makes his videos and he decides what to do. And he is not wrong

[@peterromero284]

He can do as he likes. Maybe we can eventually make language a complete impediment to understanding.

[@earyrockwell8030]

@@peterromero284 Gougu is technically and historically the name of the theorem, he's been using it in most of his vids, and he has explained it in one o his videos. Just like Bernoulli, should've been named L'Hopital's rule after him instead of L'Hopital

[@remopellegrino8961]

@@earyrockwell8030 people learn the theory as Pythagorean and l'Hopital! If you want math to be accessible for everyone, use easy anchor points!

[@oenrn]

​@remopellegrino8961 by "people" you mean "westerners ", who are a minority of the world population.

[@electricpaper269]

Maybe this is supposed to be hard but it’s really just a trivial problem. BF = 10 and EF = 5 is given.

[@PeroG]

i did this problem in diff way, i found the S of AEB and found the EB, now i found the AG with the S of AEB *2 / EB, i found the BG with AG and AB, i did all of that for the S of ABF, now all i need to do is to find the altitude from AB in the same way i did the S of ABF *2 /10 and this is 8 so x is 2

[@adamae.7246]

I did exactly the same !

[@reetadevi7603]

While solving an Indian question, It would have been wonderful if you called, the gougu theorem as baudhayana theorem, as baudhayana was an ancient Indian mathematician who discovered this theorem independently years before Pythagoras.

[@xiaoshen194]

Presh be like : I am looking for videos which uses the *Pythagoras theorem* only!!

[@vedantshah4230]

No, in his language the Gougu theorem 😂😂

[@michaelroditis1952]

You know Pythagoras had nothing to do with the discovery or the proof of the formula. Why call it like that? I am using the formula myself. So why don't you call it Roditis theorem?

[@johnjordan3552]

@@michaelroditis1952 Using the popular word to describe a subject makes it easier to communicate

[@z4zuse]

F (and A) is also an intersection of - circle, radius 5, center E - circle, radius 10, center B

[@1505-d9r]

Please stop saying gougu theorem, don’t fight against windmills

[@williamhenry4380]

Like my teachers used to say the shorter information in a problem the harder it is - In a nutshell, this math problem totally blew my mind

[@gonzalezm244]

Linear Algebra Approach Set origin at E Reflection matrix can be represented as: PDP^-1 = [ 2 -1; 1 2] *[1 0; 0 -1]*[2 -1; 1 2]^-1 = 1/5 * [3 4; 4 -3] Apply this transformation to vector [0;5] by multiplying [3 4; 4 -3] * [0; 5] = [4; -3] We take distance from y=-5 (bottom of square) to y=-3 (from reflected vector) to be the result 2

[@vinisherdaotaku3241]

Dam I was trying so hard to think outside of the box

[@Om-id1qr]

That's the mistake even I do at times but later realize it was so simple 😂

[@zamilhoquesiddique3249]

Nope, outside of the square.

[@Sawtooth_007]

Wow what a different approach using the gogou theorem. I always kept using the Pythagoras theorem but this is something new that is totally different from the Pythagoras theorem. Wow

[@j.fitness5701]

I used a different approach using similar triangles and Pythagorean/Gougu theorem. Draw a line segment that is parallel to DC and passes through point F. Let's call new point of G reside on AD line segment and H be on BC line segment. Therefore EFG triangle is similar to FBH triangle. Let's call "x" as EG distance and "y" as GF distance. Using similarity argument: 5/x = 10/FH, FH distance becomes 2x and 5/y = 10/BH BH distance becomes 2y Now we have 2 equations with 2 unknowns: 1) y + 2x = 10 (side length of square) 2) 25 = x^2 + y^2 (Pythagorean/gougu theorem) When we solve this system of equation for x, we get an expression of: x^2 -8x +15 = 0 Roots are 3 and 5 5 can't be the answer because it is the length of the hypotenuse, therefore x's value must be 3 From here we conclude that the distance must be equal to 2 since half length of square is 5 and 5 - 3 = 2.

[@WoodyC-fv9hz]

Solution is 2. Working out half the angle at the point of the paperplane (top right hand corner), I used arctan(5/10)=26.565°. Double that to get the angle at the tip of paper plane. 90° minus tip-angle gives the angle between BF and BC, (36,87°), needed to solve the problem. Since the hypotenuse is also known (10), you can solve for distance between F and the right hand side of the square via the sinus function, resulting in 6! Last step: 10 - Sqrt(10^2 - 6^2) = 2 solves the problem (Pythagoras).

[@Kosteru-des]

It is very nice that you present a lot of alternative methods! The most serious issue I face when trying to solve these problems is lack of imagination! I cannot figure out if and when I need to draw additional lines and/or shapes!

[@I_am_Abhijeet]

It can be very easily done by using equations of line assuming point D is at (0,0)

[@alexispapakonstantinou]

Thats what i did

[@aussietaipan8700]

I was able to work it out via the Trig, the first 2 were over my head. Great stuff.

[@drpkmath12345]

This was quite easy brain teaser but def useful for sure!! Haha good job!

[@ngchorhong3174]

I solved it by using the area of trapeziums to relate F to DC(x)with the two lengths that form DC (y and 10-y) Then trigonometry to find the length of y to solve for x It's so interesting that this problem has many ways to get to the solution

[@PopizzdioJazz]

Spoiler: To all of origami guys out there, this is how you divide a square sheet of paper into 5 equal parts.

[@wynchell.abanes]

I personally prefer the intersection of line EB and line C to the midpoint of AB. I know it involves two folds, but it produces a crease with a point.

[@wynchell.abanes]

But I later figured out that I should just align the edge of my paper to 10 lines of my notebook, then make small marks on the edge. Only when I don't mind a few writings on the paper.

[@trueriver1950]

What's interesting about that is that you cannot divide a line into fifths using ruler and compass. The Greeks missed out on a lot of constructions by refusing to fold their papyrus.

[@trueriver1950]

For info, with ruler and compass you can divide a line into any power of two equal parts (half, quarter, etc) and then multiply that by any integer (to get 3/4, 5/8ths etc). But using only a straight edge and compass you cannot split a line segment into n equal parts if n is not an integer power of two.

[@wynchell.abanes]

@@trueriver1950 if I'm not mistaken, using straight edges and compasses will not allow you to divide an angle to 3 equal parts. But is very doable with origami.

[@youngstervideo]

Draw a line through the point F parallel to DC, intersecting with AD at the point G and with BC at the point H. This makes two similar triangles EFG and BFH. Then we can easily find out EG=3, GF=4, FH=6, and HB=8 through the identical ratio of the corresponding sides of these two similar triangles.

[@mr.coconut2310]

I love the use of various solutions, so creative!

[@boguslawszostak1784]

Since you like different methods: Here's a method that allows you to solve this problem in your mind by using complex numbers and their basic properties. v = (2 + i) ^ 2; k * | v | = 10; x = 10 - Im (k * v) so v = 3+4i; | v | = 5; k = 2; x = 10- Im (k * (3+4i)) = 10-2 * 4 = 10-8=2 Explanation: We choose the Re axis from Point B towards Point A, the Im axis from Point B towards point C. The number 10 corresponds to the vector BA, the number 5i to the vector AE, the number 10 + 5i = 5 * (2 + i) to the vector BE. The multiplication by m * (2 + i) corresponds to the rotation of the multiplied vector by the angle ABE and the scaling with the coefficient | m * (2 + i) | If we multiply the result once more by m * (2 + i) we get a rotation by the angle 2 * ABE. Now it is enough to rescale the obtained vector so that its length is 10. For easy counting in the mind, choose m = 1. Multiplying the unit vector of the axis Re by the vector v rotates it in the right direction, the coefficient k determines an additional scale so that it has a length of 10. v = (2 + i) ^ 2=4+2*2*i-1=3+4i |v|^2=3*3+4*4=5*5 so |v|=5 we neet to multiplicateit by 2 to have lenght = 10 2*v=6+8i

[@anujsangwan6718]

THERE IS A MUCH SIMPLER SOLUTION TO THIS PROBLEM : Connect F to DC meeting it at let say G. Also conncet F to AD meeting it at let say H and also connect F to BC meeting at let say J. Now observe two triangles EHF and BJF . EF is 5 cm which a hypotenuse to triangle EHF. And BF is hypotenuse to triangle BFJ which is 10cm. Now these can be two standard right Angled triangles with sides {3,4,5}(EHF) and {6,8,10}(BJF). So let's say EH is 3 cm then HF is 4 and then the lengh HD is 2cm. HD = FG. Therefore FG = 2.

[@DeepBhatia95]

In your solution, your assumption about the side values of the two right triangles EHF and BJF is incorrect. BJF can very well be {5, 5√3, 10}(30-60-90 triangle) or {5√2, 5√2, 10}(45-45-90 triangle) or some other combination. In general, out of three side lengths and three angle measures you can know the exact values for all 6 only if you know at least 3 of them including one side. In the given question, we know one angle measure (90) and one side length (10) which is not sufficient to calculate other 2 angles and other 2 side lengths

[@SouthernHerdsman]

1. draw line extension from point F parallel to DC, intersecting ED at D' (D prime) and intersecting BC at C' (C prime); 2. let the distance between D'F be x, and the distance between FC' be y; 3. A set of parallel equations can thus be established as follows: [i] D'C' = x+y [i] x+y = 10 [ii] x/EF = (FB^2-y^2)^(1/2)/FB [ii] x/5 = (10^2-y^2)^(1/2)/10 4. solving equations for x: 2x = (10^2-y^2)^(1/2) 2x = (10^2-(10-x)^2)^(1/2) 2x = (100-(100-20x+x^2))^(1/2) 2x = (100-100+20x-x^2)^(1/2) 2x = (20x-x^2)^(1/2) 4x^2=20x-x^2 5x^2=20x x=4 The length of ED' is 3 since EFD' is a classical 5-4-3 triangle; distance from F to the line segment DC = D'D = ED - ED' = 5-3 = 2 .

[@siamsami4115]

I solved it in a much easier way. The ans is, distance = 5*(1-cos(180-2*arctan(2))) Draw a orthogonal line projection of the required distance on ED. Let's call this orthogonal line projection GD. Here, GD equals the required distance. Connect F and G. We just have to solve for GD now. Since AE=EF and BF=AB, we can say from symmetry that angle AEB is equal to angle BEF. Angle BEF = Angle AEB = arctan(10/5) = arctan(2) Angle DEF = Angle GEF = 180 - Angle BEF - Angle AEB Or, Angle GEF = 180 - 2*arctan(2) In triangle GEF, angle GEF is 90 degrees and EF is the hypotenuse whose length is 5 unit. So, EG=5*cos(angle GEF) = 5*cos(180 -2*arctan(2)) GD = ED - EG = 5 - 5*cos(180-2*arctan(2)) = 5*(1-cos(180-2*arctan(2))) = 2

[@bobzarnke1706]

Another geometric solution. Draw a line through F parallel to DC, meeting the sides of the square at G and H. Then ΔEGF is similar to ΔFHB; so GF/EF = BH/BF and EG/EF = FH/BF, ie, (10-FH)/5 = (10-d)/10 and (5-d)/5 = FH/10. Eliminating FH and solving for d gives d = 2.

[@mehmetanil1906]

Only using triangle similarities answer is easy. Angles CDF = EBA = FAD. 5/10 = y/(10-x) and x/y = 5/10 then x= 2 and y=4 where x is distance perp. to DC and y is the distance perp to ED.

[@laoamao]

Draw a horizontal line passing F. The two right triangles formed with hypothenuses 5 and 10 are similar. Let the vertical leg of the left triangle be a and the horizontal leg be b. Then the vertical leg of the right triangle is a+5 and the horizontal leg 10-b. Set up proportions of the two triangles and solve.

[@AcaPele007]

Hey Presh, I solved this one in 5 seconds, without pen and paper. You folded square to the point E, which means that segment ED and segment EF are both equal to 5. Now mark point H just at the same place where you did. Right triangle EFH has hipotenuse equal to 5,which means that it is a 3,4,5 right triangle. Graphically it is obvious that segment FH is longer than EH, thus EH =3,and FH =4. HD =ED - EH =5-3=2.

[@kevinmorgan2317]

"Right triangle EFH has hipotenuse equal to 5,which means that it is a 3,4,5 right triangle." Why must it be 3,4,5? There are an infinity of rt angle triangles with hypotenuse = 5.

[@joshuamoyer4141]

I got 2, but my method was to realize that triangle ABE and triangle FBE were mirrors of each other. This means angle AEB and angle FEB are the same, and solve for angle DEF because they add up to 180. I drew a horizontal line from point F to line Segment DE, called that point G, and solved for the length of EG. The distance from F to DC is just 5- length EG. It also works out that EFG is a 345 special right triangle.

[@kamaljethwani6363]

There is an another solution. Consider 2nd end point of green line (0:23) as 'G'. So we have to find the length of line FG. Now figure is divided into four parts. Two right triangle and two trapezium. Trapezium 'DGFE' and 'CBFG'. Sum of area of both trapezium= ->Area if square - 2*Area of triangle(AEB) ->100 - 2*1/2*5*10 ->100-50 ->50.….......(1) Area(DGFE)+Area(CBFG) = 1/2*DG*(DE+GF)+1/2*GC*(BC+GF) ->1/2*DG*(5+GF)+1/2*(10-DG)*(10+GF).......(2) Equating (1) and (2).. 1/2*DG*(5+GF)+1/2*(10-DG)*(10+GF)=50 ->DG*(5+GF)+(10-DG)*(10+GF)=100 .......….........(3) Now assume a line segment starting from 'F' and parallel to 'DC'. Which cut 'ED' at 'H'. Now 'EFH' is a right angle triangle. With hypotenuse (EF) as 5. Since EF is 5 so the value of EH and HF can only be either 3 or 4 for. Means, EH=3or4 HF=3or4 Since HF=DG, Put value of DG(first 3 and then 4) in equation (3). Let, DG(or HF)=3 (or EH=4) ->3*(5+GF)+(10-3)*(10+GF)=100 ->15+3GF+70+7GF=100 ->85+10GF=100 ->GF=1.5 Since, GF=HD and, EH+HD=5 So, EH=3.5 (Which violate our assumption) Now let, DG=4 ->4(5+GF)+(10-4)*(10+GF) ->20+4GF+60+6GF=100 ->80+10GF=100 ->GF=2.......Answer.

[@T3sl4]

I did this by slopes. Which is basically the trig form if you like complex numbers! Note that EB has slope 1/2. FB has twice the angle. We can find the slope by remembering that complex numbers rotate when multiplied together: thus (2 + 1i)^2 = (4 - 1) + (2 + 2)i = 3 + 4i. Thus FB's slope is 4/3. If we construct the projections of F onto AB and BC, we get P (@5:35), and let's call the other Q (similar to H @2:25, but to the right). BQ/PB = 4/3, and they form a right triangle whose hypotenuse is 10, so we have twice a 3-4-5 triangle: PB is 6 and BQ is 8. Thus QC is 10 - 8 = 2, the same as the distance from F to CD.

[@pascaldelcombel7564]

Please stop with Gougou. It doesnt matter how many speak this or those languages: my name is Pascal and not other language can change it to something else. Pythagore does not spell G O U G O U.

[@cr1216]

There is a really simple way: Draw perpendicular line from F to ED intersecting at X and F to BC intersecting at Y. We know triangles EXF and FYB are similar with ratio 2 because the hypotenuses have lengths 5 and 10. Now let EX=a and XF=b, then FY=2a and YB=2b. Now we have a system of equations (b+2a=10, 2b-a=5) solve it and a=3, b=4. Therefore distance from F to CD is 5-3=2.

[@cr1216]

No calculation required beyond integers.

[@eleonorav.d.d.8864]

How I did it, name the distance from F to DC x and the distance from F to BC y. Then you use the Pythagorean Theorem to get two formulas with two unknown. (10-x)^2+y^2=10^2 and (5-x)^2+(10-y)^2=5^2 you write out the squares and subtract the one from the other to get rid of the x^2 and y^2. You find the relation x=2y-10. Entering this back in the first formula gives a kwadratic equation for y. Solving it gives y=10 or y=6. Taking y=6 gives x=2 because of the earlier relation. So you can do it with only using our friend Pythagoreas

[@earyrockwell8030]

I GOT ANOTHER WAY. I put two segments DF and FC creating a new triangle DFC and simply look for its area by subtracting all the areas of other triangles from the area of the sqr = 100, ABE is obviously 25 and same for EBF; DEF can be found by Area=(0.5)(ED)(EF)sin(theta), *theta BETWEEN them* This theta can be found by trigonometry and the same way for finding the area of FBC. Area of quadrilateral AEFB = 50 triangle DEF = 10 triangle FBC = 30 Thus; Area of FDC= 10 With known base 10, its height should be 2.

[@pbierre]

If you're comfortable using direction vectors and angles interchangeably, it's straightforward to compute the coordinates of the folded-over corner F. dirVec(E --> B) = [2/sqrt5, 1/sqrt5] dirVec( F-->B ) = dirVec( 2 * radiansOf(dirVec(E --> B))) = [3/5, 4/5 ] (folding doubles the angle) F = scalarMult (dirVec( F-->B ), -10 ) = [ -6, -8 ] (origin at B) F.y = -8 (2 units above the lower edge of square at y = -10

[@ItachiUchiha-xb9kb]

compute area of square = area of( 2 congruent triangles + area of smaller trapezium + area of bigger trapezium). Let the point where F touches DC be M. Take FM=x and DM=y. by above equation we get relation between x and y i.e., y=2x. now by trigonometry find the angle DEF. angle AEB = arc tan(2) = angle FEB(by trigonometry). and angle DEF turns out to be (180 - 2*arc tan(2) ). draw perpendicular from F to DE meets at H. we get a right angled triangle EHF, where we got angle HEF and one side is known i.e., hypotenuse is 5 units. again use trigonometry to find value of HF. which turns out to be 4. so y=4, by using above equation we get x=2. and we are done

[@lakshaygarg1544]

I loved the method with trigonometry It was amazing

[@platoscatboy9772]

Right? So elegant

[@edwardswing9741]

I solved this problem another way. Form the line XY through F that is parallel to DC. Now, we know that EXF (a right triangle) and FYB are similar, and of the ratio 1:2. Now EX+XD=5 and BY+YC=10, and XD=YC. After a bit of algebra, we can easily deduce that the distance from F to DC is 2

[@ravenking2458]

Yep same

[@wesleydeng71]

I clicked this video just to see the "Gougo Theorem". And I was not disappointed! 🤣

[@劉人名]

uses the Pythagoras theorem

[@ezequielcallegari2644]

That's what he did

[@HistoricalPlayer]

Turkey's geometry education is full of this kind of questions nowadays :)

[@johnjordan3552]

Really?

[@kadir4965]

@@johnjordan3552 yeah he said true

[@tolgacln2346]

@@johnjordan3552 yup

[@piman9280]

No doubt they gobble up such questions - I'll let myself out!

[@muhammedemretekin6889]

Yes, i solved this in 10 sec.

[@zhenyuanlu1764]

Let x = angle ABE. Then sin(x) = 1/sqrt(5), cos(x) = 2/sqrt(5). sin(2x) = 2 sin(x)cos(x) = 4/5. d = 10 - 10* sin(2x) = 2

[@Kawizend]

you can just draw a parallel line to DC passes through F and then solve it using similarity

[@Tiqerboy]

That's how I did it :-) Though I did set up a quadratic formula to solve for the desired quantity, there might be an easier way without the quadratic formula.

[@RashadSaleh92]

I used that line to split the thing into areas then find the sum of the areas to be 100. It required calculating the sin of the angle fBc, which wasn’t that hard... but i am guessing your method doesnt need to?

[@Kawizend]

@@RashadSaleh92 as we know that EFB angle is 90 degrees, we can see two similar triangle there, and as we know EF is 5 and BF is 10, their rate is 1/2 and we can solve that way (we dont need trig)

[@trueriver1950]

An easier solution using similar triangles is to draw a horizontal line through F (that is draw a perpendicular to line ED) meeting ED at H. This triangle is similar to AED.

[@dwinson]

Yes and this method avoids Pythagoras/Gou-Gu

[@vishalmishra3046]

AB = 10 = BF and AE = 5 = EF and EB is common so Δ AEB is congruent to Δ FEB so BE bisects angle ABF (say each angle is T). So, Answer = 10 - PF = 10 - 10 x sin 2T where tanT = 1/2 sin 2T = 2 sinT cosT = 2 tanT / sec^2 T = 2 (1/2) / (1 + 1/4) = 4/5 Therefore, Answer = 10 - 10 x 4/5 = 10 - 8 = 2

[@JLvatron]

I was sure there was a similar triangles solution but didn't figure it out. I used trgonometry, but took a longer path than your elegant solution. Thank you!

[@restablex]

This is what I did drive.google.com/file/d/1TQY4e8k3gy3su1m77qmVMaRlq78p493s/view?usp=drivesdk

[@salamander5703]

Interesting it comes to a nice round 2! I tried a different way (which failed) but along the way found another interesting thing. Project BF until it hits CD. Connect that point to E and using similar triangles you find that the triangle BC and the new point has sides 12.5, 10 and 7.5 ie a 3,4,5 triangle!

[@derwolf7810]

Another solution would be using the same drawing as in : Triangle(AGB) is congruent (symmetrie: rotate the square by 90 degree) to triangle(DFA), so F to the bottom is the same length as G to the right. Then define H on line EA, such that triangle(EHG) is similar to triangle(EAB) and see that triangle(EGA) is also similar and that they are all sticked together with the shortest side to the second shortest side (|AE|/|AB| =5/10 = 1/2), so: 2 |EG| = |AG| = |GB|/2 |GB| == 4 |EG| => searched length = |HG| = 10/(1+4) = 2

[@brijesh4723]

Just apply Pythagoras theorem ..

[@rafazieba9982]

Find a point that is 5 from point E and 10 from point B. Those are two quadratic equations. Subtract one from the other to get a relation between the x and y coordinate. Solve for y and place it into one of the quadratic equations. you'll get a quadratic equation without a free element (x^2 - 4x = 0 => x (x - 4) = 0). x = 0 or x = 4, so y = 10 or y = 2. You have points A and F.

[@davidgillies620]

An insight from 2-D geometric algebra: rotations can be considered as reflections across the line bisecting the rotation angle, and the converse is true. So F is just A rotated about B by twice angle ABE. or 2 arctan(1/2). Make the 2-D rotation matrix for that angle ((3/5, -4/5}, (4/5, 3/5)) and apply it to the point (-10, 0). The required distance is 10 plus the y-component of this rotated point, which is 2. It helps to know your double angle formulae and trig functions of inverse trig functions, but if you're applying to do maths at university that should be a given.

[@bobzarnke1706]

In a more general case, where the square has side s and point E is ks (k < 1) from vertex A (k = 1/2 above), then the distance is: s(1 - k)^2/(1 + k^2).

[@valirezrezag4674]

Hello I have this puzzle that I heard when I was young (25 years ago) The best strategy for dividing a cake between two people is: One of the two people cut the cake and the other one chose first. In this context: What is the best way to divide a cake between three people? What is the best way to divide a cake between n? I hope everything is clear and understandable. I apologize for the English language as it is not my native language. thanks

[@eleonorav.d.d.8864]

If you want to see one approach to this problem, Numberphile did a video about it. kzbin.info/www/bejne/oZKwfHykoLtqjNU

[@valirezrezag4674]

@@eleonorav.d.d.8864 Thank you very much

[@TheQEDRoom]

If you add points H and I such that HI // DC and HI passes through point F, you can create similar triangles EFH and FBI with hypotenuse 5 and 10, respectively. Let FH=a and FI=10-a. Solving for a, we will get a=4. Double this and we get the height of the bigger triangle is 8. So F is 10-8=2 units away from DC.

[@shirshagarwal5221]

It's very easy I did it using Pythagoras theorem Consider that distance x Join F to side AD at X and BC at Y use Pythagoras theorem in triangle EXF and triangle BYF taking EX and BY in terms of x We get XF and YF in terms of x Add XF and YF and equate it to side length 10 So I got the answer x= 2

[@CAMOBAP795]

Also possible to solve with a system of 3 equations H, F, K (K on the same line as H & F and on side BC) x - what we need to find FK = y HF = z We will got a system: 1. (5-x)^2 + z^2 = 5^2 2. (10-x)^2 + y^2 = 10^2 3. y + z = 10 after simplification, we will have a kind of complex equation: (25-(5-x)^2)^0.5+(100-(10-x)^x)^0.5 = 10 but root can be simply guessed: 2

[@eleonorav.d.d.8864]

If you rewrite 3 to the form z=.... and substitute it in 1. Now write out both 1 and 2 and subtraction 2 from 1 (this way you lose the square variables) this new equation is a simple linear relation between x and y. Enter this in equation 2 to get a quadratic equation with one variable, which is easily solvable

[@gokuvegeta9500]

@@eleonorav.d.d.8864 How do we know if both circles intersect at point F ?

[@thaissacysne6509]

I did it in a different way... First i draw a parallel to AB, crossing the point F. Call the new point P (crossing the AD) and Q (crossing BC) Then I discovered that angle PEF = angle BFQ and angle EFP = angle FBQ Call the distances EP = x, BQ = 5+x, PF = y, FQ = 10-b Using the propotion between the triangles PEF and FBQ, I have that 5+x = 2y and 10-y = 2x (And we need to have in mind that it makes x=2y-5 and y=10-2x). But I need the value of the perpendicular line from point F to line CD, but this value is 10-2y. That makes x = 5 - (10-2y). So, x = 2y-5. As y = 10-2x, it means x = 2(10-2x)-5 = 20-4x-5 = 15-4x. And if x=15-4x, then x=3 Finally i discover that the distance between point F and line CD is 5-3 =2

[@jasonrs8141]

I solved it by drawing a parallel to the segment that passes through AB so that it passes through point F, then we notice that there are 2 similar triangles in the ratio 1: 2, so we have that if we call x and k the sides of the minor triangle, we obtain that the segment parallel to AB is equal to 2x + k = 10 and also by the Pythagorean theorem we have that x ^ 2 + k ^ 2 = 5 ^ 2 then we obtain two solutions: k = 1 and x = 0 or k = 4 and x = 3 of Here we discard the first solution since x is one of the sides of the triangle, finally as x = 3 and the segment ED measures 5, so the value of F is equal to 5-x, that is, 5-3 = 2

[@pkd2762]

I did the same way! This needs only one construction and to me looks to be a simpler solution.

[@rafaelliman8167]

I did the same way until the 2 similar triangles with the ratio 1:2, but then I'm simply using the legs of those triangles. Suppose the distance we're looking for is a, which can be projected to the vertical sides of the square. Because those 2 triangles are similar with the ratio 1:2, we'll find that x=5-a & 2k=10-a --> k=5-a/2. Simply substitute these two to the fact that, as you say, 2x+k=10, and we'll find a, which is what we're looking for. This is even MORE elegant, imo, since we don't even NEED Pythagoras (or, according to Presh, 'Gougu'...)

[@Adoom-tk8bq]

I solved it in yet another way. First, we know triangles EAB and EFB are similar. Solve for their sides and angles (sides 5, 10, and 5*sqrt(5), angles 90, 63.43, and 26.57). Next, draw a horizontal line from point F to the line BC. Call the intersecting point G to form a new triangle: FGB. Solve for the angle FBG by subtracting 2*26.57 (solved from the triangles above) from 90 to get 36.86. The hypotenuse is line BF which is already known to be 10. Using SOHCAHTOA, we know cos(36.86) = BG / 10. Solve for BG: BG = 10 * cos(36.86) = 8.00. Last, the distance between F and line DC is 10 - BG --> 10 - 8 = 2.00.

[@michaeldakin1474]

A slightly different solution, that draws upon both similar triangles and trigonometry... Create triangle DFG, with G being the point at which the vertical line from F intersects with DC. Note that DF || EB (for proof, calculate angle ABE (hint: arctan 0.5), from which you can readily calculate angles AEB, BEF and DEF. Note that DEF is an isosceles triangle, so the bisector of angle DEF will be perpendicular to DF, and half of angle EDF = angle FDG, which so happens to be also equal to angle ABE). So, triangle DFG is similar to triangle ABE. Therefore FG = DF/(5^0.5). To find the length DF, calculate sine(half angle DEF) and multiply by 10 (being hypotenuse of 5, and then double because the calculation gives just half of DF).

[@jihyepark9139]

Wow, even you just fold a paper, you still can see Math in it. 🤣😂 Math is everywhere.

[@qingyangzhang887]

I did it using a different method that is arguably the dumber method because it involves less thinking and more algebra: 1. When you fold the edge over, you create 4 regions, two identical triangles and 2 trapezia 2. Area of square = sum of these areas The height of the two trapezia can be figured out I get to a quadratic, which leads me to one extraneous solution and 1 solution = 2.

[@phos4us]

A line from point D to halfway between B and C makes the line y=1/2x and passes through F. A line from point A to halfway between D and C makes the line y=s-2x (s = side length) and passes through F. Find the y value of their point of intersection (F) and you get s/5. A fifth of the side length. This way needs proof that F exists on the lines but I know from something else I've done it does. edit: He did it a better way in the video in terms of not knowing if they intercept beforehand.

[@ILKERUMAM]

Algebraic form of solutions using X and Y also gives the same answers, Though trigonometric approach seems to be the best way accord to me

[@apratimtewari4288]

yes. I solved it by defining 2 variables. After that, it comes down to the Pythagorean theorem and simple algebra

[@The_Curious_Ra.ke01]

I solved it in a second without any calculation. I'm Origami artist and I know when fold a paper like this distance between F and DC will be 1/5 of the side of paper.

[@vectorclassic6403]

u r too slow boomer i solved it in 0.00001 milli seconds

[@brahmandsaraswat867]

The third part is the best way I found to solve this problem. Their is one more way, but it requires the essence of trigonometric inverse functions, so leave it.

[@yassinesafraoui]

a much simpler solution can be found with trig functions, we can find the angle AE^AB( it's arctan(2)), and we can see that the angle between AB^EF is equal to AE^AB, thus we can deduce the angle DE^EF( pi - 2*arctan(2)), and EF is equal to AE( which is 5), we can deduce the distance that's the projection of EF on ED( 5 * cos( pi - 2* arctan(2))), and EF is simply 5 minus that distance

[@Queenside_Rook]

I also used trigonometry but went an entirely different direction. Let's call the distance in question y. Angle BEF = tan^-1(FD/EF) = tan^-1(2) Angle AEB = angle BEF Angle DEF = pi - angle AEB - angle BEF Consider a point, G, that lies on line AD and is the same distance from point D as point F is from line DC Line GD = 5 - line EG cos(angle DEF) = line EG / 5 Line EG = 5cos(angle DEF) Line GD = 5 - 5cos(pi - 2tan^-1(2)) Line GD = 2 y = line GD = 2

[@chhabisarkar9057]

Dear presh sir , i think you've chosen the more complicated methods , just use simple Pythagoras and it'll get solve in seconds (never knew such a problem would come from CMI , one of the most reputated mathematical institute in india ) . A better method can be this - Draw a horizontal like from the top of the segment to 2 sides , on the right side (DC) let bigger part be x and other part will be 10-x , on the right side (AB) for the smaller triangle the perpendicular will be 5-x and remaining smaller part at the bottom will be 10-x . use *PYTHAGORAS* and you'll get the horizontal line = √25-(x-5)^2 + √100-x^2 = 10 , notice x= 8 is a solution (or just solve it ur self) So the segment = 10-8 = 2 😁 Edit - that trigonometry method was neat tho but I still prefer plane geometry

[@dezenterrier]

Let G be projection of F to DC. Then DGFE and GCBF are trapezoids. We know sum of their area, which is area of square minus 2 areas of triangle ABE. Also we know EF and FB. We have 2 unknowns: FG and DG and 2 equations: first one comes from area and the second one from some Pithagoras EF^2=DG^2+(ED-FG)^2. We solve it and get the answer. We can also use similiar pitagoras equation from trapezoid GCBF.

[@Escviitash]

G=Heigth from A in ABE= (5x10)/(sqrt(5^2+10^2)) AF=2xG H=Angle ABE= arctan(5/10) I=Angle BAF = 90-H K=Vertical distance AF= sin(I)xAF Solution=10-K

[@JohnDlugosz]

I used a method similar to the third, but without needing any actual trigonometry (that is, knowing all the identities). Rather, a hybrid between geometry and a knowlege of how sin/cos/tan work in a right triangle. Don't bother computing the length of the hypotenuse. Just know that the tan(θ) is 5/10, so get the ATAN on my calculator. The folded triangle is similar to the vacated triangle, so the angle formed by the flap at E is twice that of AEB. Now I realized I found the wrong angle to start with, but want its complement. Double that, and the part remaining, DEF, is the suppliment of that. Now I want the projection of EF onto the DE side, so take the COS of that angle. The "drop" EF is 3, I started at 5, so the distance remaining is 2.

[@Vibaravi]

Let the projection of F on DC be point X. Let FX be x and XC be y. Form 2 simultaneous eqn and solve for x.

[@harryxiro]

The way I solved it was by working out the angle AEB by doing the inverse tangent of 10/5. I then multiplied that angle by 2 because the angle BEF was the same angle as AEB because the angle was reflected, and I did this to get the angle AEF. I then subtracted AEF from 180 to get angle FED. I then did some algebraic manipulation of Cos = A/H and turned it into H x Cos = A to work out the adjacent. FE was length 5 because it was translated from AE so I did 5 x cos(FED). This got me the answer of 3, and because AE was equal to 5 and BC was equal to 10, that means ED is also equal to 5, so I then subtracted 3 from ED to get me the answer of 2!

[@יהלקמינר]

I solved it with the angles. I found AEB that equals to DEF (63.43) with trigonometry that means that DEF=54.14 and then trigonometry again

[@ericfuchs123]

I just used the right triangles EHF and BFQ (Q being a point between C and B) to solve for the unknown distances HF (X) and HD (Y) Use Grogu Theorem. We know the two hypotenuses are 5 and 10. Two equations are: (5-Y)^2 + Y^2 = 5^2 (10-X)^2 + (10-Y)^2 = 100 Then do algebra. You’ll find that X^2 + Y^2 = 10X in the first equation, plug that into the second. You’ll get X = 10 - 2Y. Plug back into the first. Y = 4. X = 2, so HD is 2, and that’s the answer.

[@asavariparanjape3556]

Easier solution : Name the point where the Height touches CD as O Now draw a line parallel to EF passing through point D Say line FO and line parallel to EF from D meet at point G Now let FO be h Then GO will be 5-h Also EF =5 and BF=10 (as the part is folded) So EF = DG (properties of parallelogram) So DG =5 Using Pythagoras thereom calculate DO extend the line FO upwards and let it meet AB at X So AX will be equal to DO Now substract to get XB and OC. Now calculate area of square and substract it from the area of triangle ABE and EFB. The remaining area is the sum of two trapeziums whose area can be calculated. Ultimately it ends up being a quadratic equation which is a easy to solve. And you get the ans to be two

[@samehhassan9066]

I liked the coordinate solution,it is out of the box.

[@deprivedoftrance]

Since the paper was folded, then EF=5 and BF=10. Drawing a parallel line HK passing through F, you have a right triangle EHF with hypotenuse 5. Clearly a 3-4-5 triangle, giving DH=2 and thus the distance DC to F as well.

[@ajbs_alexo_713]

I personally calculated the angle ABE (which is tan^-1(1/2)) then EBF (is equal to ABE), afterwards I took the coords of A in the (B, BA, BC) plan, rotated A by an angle of ABE + EBF with center B (using the rotation matrix) to find the coords of F (0.6, 0.8). Finally, I took the difference 1 - 0.8 = 0.2 (1 is the 2nd coord of C following BC) and multiplied it by 10 since BA and BC have a length of 10.

[@alfredkhew1634]

I solved it with trapezium. Say the length we are looking for is h, mark it FF' which is perpendicular to DC. Let DF' be x. We have 2 right angle triangle (ABE and FBE) and 2 trapezium (AFF'D and BFF'C). Add them up equals to the area of the square 10^2, with some algebra we get h=x/2. Say FG is perpendicular to AD, observe that the small triangle EFG which has sides (5, x, 5-h), using Pythagoras theorem we'll get x=4 which in turns h=4/2=2.

[@leonhardeuler3127]

I used the fact that triangle DEF and BFC are isosceles triangels. The area of DEF+BFC+DFC= 50. Name the angle E in DEF E1 and the angle E in ABE E2. The sine of E1= sin( 2*E2)= 4/5 Now you can use this formula for the area of DEF:1/2 * 5^2* sin(E1)=10. We can do the same steps for BFC but sin(B1)= cos( 2*B2)=3/5 . With the same formula we find for the area of BFC:30 . This means that the area of DFC has to be 10. If we solve the equation 10= (10*x)/2 we get that the distance F is equal to 2. * In fact we can do these exercise faster. Since we know sin(2*E2), you have to notice that sin(2*E2)=sin(2*B2)=4/5. The cosine of B1(=FBC) is then also 4/5. now you have to construct triangle BFS ( S= right triangle) BS/BF= cos( B1)= sin(2*B2)=4/5. BF=10, so BS=8. Distance F= 10-8=2

[@RadekBuczkowski-h2y]

There is another trigonometry method: AB = FB = 10 AE = 5 EB = √(10² + 5²) = 5√5 sin(∠ABE) = AE/EB = 1/√5 Now place a point X between B and C on the same height as F, so that the triangle △FBX is a right triangle. This way the distance we are looking for can be expressed as following, so we only need to find BX: x = 10 - BX ∠FBX = 90° - 2(∠ABE) FX/FB = sin(∠FBX) = sin(90° - 2(∠ABE)) From the trigonometric identities we know that: sin(90° - 2θ) = cos(2θ) = 1 - 2sin²(θ) FX/FB = 1 - 2sin²(∠ABE) sin(∠ABE) = 1/√5 FX/10 = 1 - 2(1/√5)² FX/10 = 1 - 2/5 FX/10 = 3/5 FX = 6 FX² + BX² = 10² BX = √(10² - 6²) = √64 = 8 x = 10 - BX x = 2

[@RohitKulan]

We can use the point F to fold paper into a 5x5 grid, useful if anybody wants to fold Jo Nakashima's seamless cube.

[@MaxMathGames]

Very soon, we will be calling Pythagoras theorem as Presh Talkwalkar theorem. Well explained 👍👍👍