How To Solve The Shape-Shifting Triangle

This delightful problem comes from a German math competition. What's the perimeter of the triangle CDE? You have 3 minutes to solve this problem. Watch the video for a solution, and to see the shape-shifting triangle hidden in this problem.
Sources
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www.mathe-kaenguru.de/wettbewe...
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[@arnolfhufano4771]

my mind is DE+AD

[@cristiluchian3567]

Lol

[@rthomas2]

So your mind is obtuse?

[@garryock2530]

I cant solve your mind equasion too

[@wobblyorbee279]

@@rthomas2 I think his mind is two unrelated connected lines

[@TheOriginalFayari]

Why did you give such an unnecessarily complicated explanation when there's a much simpler solution? AB + DE = AD + BE 3 + DE = 6 - CD + 5 - CE DE + CD + CE = 6 + 5 - 3 = 8

[@mahmoudsalla9488]

TheOriginalFayari I was about to write this answer It's easier and faster

[@Karim-qv5gw]

Me too

[@tejjj3]

hey, why are those 2 sides equal?

[@user-vj7uc9tj7c]

Tej DIvadkar In a circumscribed quadrilateral, the two pairs of opposite sides add up to the same length. Which means: AB + DE = AD + BE.

[@owentate502]

If I instead received your explanation, I would have learned and understood nothing.

[@chessandmathguy]

Explanation was elegant and perfect. Animations were on point. Made perfect sense. Nicely done sir! Thanks for posting.

[@ceulgai2817]

I have to say, despite people going on about how some of your videos are "too easy," even on vids where I should be able to quickly solve the problems, there are principals that I've forgotten about. As such, thanks for the reminder!

[@adamstanisawski1980]

The problem becomes trivial when you use Pitot theorem ("in a tangential quadrilateral (i.e. one in which a circle can be inscribed) the two sums of lengths of opposite sides are the same" - from Wikipedia). That means that AB + DE = BE + AD and therefore DE = BE + AD - AB. The permiter of desired triangle is EC + DC + ED = EC + DC + (BE + AD - AB). Since BE + EC = BC = 5 and AD + DC = AC = 6, perimeter equals BC + AC - AB = 5 + 6 - 3 = 8. QED

[@atlas7425]

Actually, "QED" is an unelegant ending in this case. You didn't want to proof a specific statement to be true or false, you wanted to to calculate a fixed value. It could refer to the statement that the problem became trivial when someone used the Pitot theorem, in which case you've only shown that the problem is *solvable* by using Pitot theorem, though.

[@professorpaulo1120]

Adam Stanisławski, wonderful!

[@cmarley314]

Much less confusing than the video (which I skimmed through without sound), but how could one prove that quadrilateral property thing..?

[@Skandalos]

Christopher Marley, the proof is really simple, just make a sketch, name the tangents and set up the sums of opposing sides using these names.

[@adamstanisawski1980]

Mac KNÖDEL You're right, I shouldn't use QED because I'm not proofing the length of the permiter but calculating it. My mistake.

[@ichigo_nyanko]

Every time you make one of these videos I think I've found the most ingenious way to solve it before playing the video and realizing you did it the exact same way.

[@stevethecatcouch6532]

The perimeter is 8. There are a variety of ways to see that the perimeter of CDE is the sum of the lengths of a tangents from C. My favorite is to note that because the positions of D and E are irrelevant, I can put D on the point of tangentcy on AC and put E on C. The perimeter of ABC is 14. The combined lengths of the other four tangents is twice AB, 6. So the two tangents from C have a combined length of 8.

[@jimlocke9320]

The perimeter of triangle CDE is the sum of the lengths of CD, DE and CE. DE's length is the sum of the lengths of DJ and EJ. So, triangle CDE's perimeter is the sum of lengths (CD + DJ) and (CE + EJ). However, DJ and DF are equal in length, so length (CD + DJ) = length (CD + DF) = length CF. EJ and EG are equal in length so length (CE + EJ) = length (CE + EG) = length CG. So, triangle CDE has perimeter equal to the sum of lengths CF and CG, which Steve proved is equal to 8.

[@adanartan3425]

Thank you for your informative knowledge. It’s amazing to watch your videos. Please keep doing more videos!

[@johncena-jh3of]

Is it possible to get the question papers for this particular contest?

[@omrigal8]

I think that solving this problem using the circle inscribed in a quadrilateral formula is way simpler. Apply the formula on ABDE: DE + 3 = AD + BE Since AD and BE are equal to AC and BC minus the lengths of DC and EC you can solve the perimeter of triangle DCE: DE + DC + EC = 8

[@harshitverma685]

Me 2

[@davidz8744]

What I did was to split AB into x and 3-x and created the points F and G where the circle is tangent to sides AC and BC. I then connected the points where the circle touches the sides AB, BE, DE, and AD to the center of the circle and saw that the length of GC is AC - x = 6 - x. Then we know that FC is BC - (3 - x) = 2+x. If we name the point where the circle is tangent to DE as Z, we see that FE = EZ and GD = DZ. In conclusion, we can say that EC + ED + DC = CG + CF= 6 - x + 2 + x = 8.

[@trelligan42]

Excellent, one of the best of your recent videos. This is a truly great problem, and I learned a useful principle.

[@Mike-rw1jw]

JFYI, this is elementary math meant for high school kids who are expected to solve it in 3 minutes

[@YourPalAlRetroGamer]

Presh, you didn't need the Gougu Theorem at 2:28. The RHS Condition of Congruence takes care of that.

[@jiaming5269]

If you're rushing for time, notice that the question implies the perimeter would stay the same not matter where the tangent is, use that and so place that tangent very close to F, and you'll find that the perimeter "should be" 2CF.

[@BigDBrian]

Yeah, that's how I started off.

[@LittleYusuf]

If someones rushing how would they know what *F* is

[@jiaming5269]

I dont get it.. The name is obviously arbitrary, i just referred to it as F because the fresh toadwalker did.

[@derrickthewhite1]

Yep, this is how I did it. You still have to figure out the length of F, but that's MUCH simpler than the algebra shown.

[@Zollaho]

Yeah : what I did too, from the start. 1 minute to draw the thing on paper and give a name to the the distances (p ...). 2 minutes to remember the equal distances ( the inscribed circle of a triangle has it's center at the intersection of the bisectors, hence the right equations pop up immediately with no need to demonstrate anything), 3 minutes to write and solve the equations then calculate 2CF. 6 minutes. No rush, the lazy way !

[@sanseng000]

I see some users are abusing the author's name, the way he pronounces and whatnot. I wonder are you math lovers? And if yes, why are you not focused on the math part. It is evident that it takes a lot of passion and effort to keep making these videos. This effort demands respect, and if not respect consideration at least, ridiculing unnecessarily is not helping anybody.

[@PeterJavi]

You're completely right you know. This is after all a math channel and the viewers should be focused on the math. Trash Moatwalker, you have my utmost respect.

[@shortcatz]

I totally respec Pest Thotstalker what do you mean, he get my views anyway because im an insecure adult who wanna know if im smarter than a chinese 5 grader which is obviously true cause i can solve this problem faster due to me having an iq of 369 and im undoubtedly that 0,01%

[@franciscoabusleme9085]

Im dying

[@custardtart1312]

Who made you King dear sir? One can enjoy the maths (not math - another grating Americanism), yet find the delivery style incredibly grating and irritating. The speaker has an over-excitable style, verging on a Kermit the Frog voice. The content, however, is excellent and he should be commended on this.

[@Ni999]

San Seng You've never turned on closed captions for Presto Walkers work.

[@devastated6180]

The first ever question in this channel that I managed to SOLVE

[@UPBoardPCM]

By which application you make video please reply

[@jbtechcon7434]

This problem and the presented solution is great illustration of where good test-taking strategy differs from sound mathematical methods. We are not told and have not proven that the solution to this problem is independent of specifically how segment DE is situated in there-but the problem isn’t solvable unless that’s true, so just assume it’s true and solve the problem! You’ll get either the right answer or a wrong answer that’s the fault of the test-maker.

[@Spongman]

if DE isn't specified, then either the question is broken, or the solution doesn't depend on it. the first thing that popped into my head when hearing the question was 'oh, DE isn't located, what if i just make it flat?'

[@jbtechcon7434]

+Spongman Exactly, and well said.

[@kresimirnezmah5393]

His solution does not show it, but it can be solved without assuming anything about DE.

[@JackRule16]

krešimir nežmah I'm pretty sure all they're saying is that's not very helpful in a timed test setting

[@rmsgrey]

Actually, he proves that the angle of DE is irrelevant with his initial approach, then goes back to explain the reasoning that assumes the question is well-formed.

[@ZahRa.H.]

Thank you, Presh, your solution gave it to me new perspective to the problem, but I did it by using similarity of 2 triangles. By Heron formula we can find area of ABC and then by two other formulas (A=ABxh/2 and A=RxP/2) we can find r and h of ABC. Then with subtraction (h(ABC)-2r(ABC)) we find h of CDE. Then find p of CDE bu equation h(ABC)/h(CDE)=P(ABC)/P(CDE)

[@jameswilson8270]

Great choice for a problem and great quality as always! +1

[@user-mx8sj1nc6v]

I solved it like Tehom who responded here, in even fewer steps, but I wouls like to mention that I had in mind to solve it like this, if no other way works. Find the area of ABC with heron. Find r with r=S/p , then assume DE||AB and use tales. (similarity).

[@jonni2734]

I used the fact that the figure ABED has the property AB + ED = BE + AD Then I wrote the formula of the perimeter: pCDE = DE + EC + DC = DE + BC - BE + AC - AD Then from the first equation I wrote DE - BE - AD = - AB and I substituted it in the second equation: pCDE = DE - BE - AD + BC + AC = -AB + BC + AC = -3 + 5 + 6 = 8

[@aspirant2480]

This question was also asked in NMTC exam of India. Nicely explained sir 👍👍

[@z000ey]

Idk if I made mental mistakes, but I came to the solution much quicker. First, I devolved the limit position of the triangle as you did, thus the perimeter is 2*CF or 2*CG, as you wish. But then I labeled each length of the primary triangle segments that are tangent to the circle as AF=AH=a, BH=BG=b and CF=CG=c, and just put them in the simplest equation of: a+c=6 a+b=3 b+c=5 Solving for c you get c=4 and the perimeter of the smaller triangle is thus 2*c=8

[@bend.manevitz8261]

I didn't read all 400-plus comments, so if this is up there, sorry. I don't think you needed to go to the whole, "move F to C" idea. (Using the image at abt 3:46, but adding point H between A and B...) Perimeter CDE = Perimeter ABC - (AB) - (AD) - (BE) + x + y AD = y + AH BE = x + BH and note that AB = AH + BH Substitute and simplify (I'm rearranging as I go): = 3+5+6 - (3) - y + y - AH - x + x - BH = 3+5+6 - 3 - (AH + BH) = 3+5+6 - 3 - 3

[@putindavid1911]

Took literally 1 min to solve for me being a 10th grader in India. Easiest problem I have seen in this channel.

[@EssentialsOfMath]

Let DE be tangent to the circle at point X. Call the tangency points of the incircle with the triangle L,M,N, with L on AB, M on BC, N on CA. Then DN = DX and EM = EX. (Tangents to circles from a point are equal) Thus DE = DN + EM. Then the perimeter of CDE is CM + CN = 2CM, because CM = CN (again, equal tangents.) Suppose CM = CN = x, BL = BM = y, and AL = AN = z. Then x+y = 5, y+z = 3, z+x = 6. Solving this system we find that x=4, so the perimeter is 8.

[@problematicpuzzlechannel6663]

Got this one!!

[@viniciusfernandes2303]

Thanks for the video!!!

[@NestorAbad]

Oh by the way, the 30 problems have to be solved in 75 minutes, not 90. At least here in Catalunya. I think the rules are the same across all the countries that hold this contest.

[@MindYourDecisions]

Thanks! I said 90 minutes from the Math StackExchange post, but good to know it's actually only 75 minutes--makes it even an harder time constraint!

[@-Apostolos]

In Greece it's also 90 minutes long.

[@user-mx8sj1nc6v]

Teachers who know very well math sometimes forget that it took years to develop this skill. I hope more time is given to tests like this. On the other end it is my understanding that in AMC for example you should solve about a half of the questions - to be in a good shape.

[@anandsuralkar2947]

We have jee exam in india where 330questions havento be solved in 90min only the level is very very hard.not like this easy one

[@BuffAssassins]

we have this exam, 5th graders have to solve 990 calculus problems in 5 minutes

[@THEGANJASAINT]

THANK YOU VERY MUCH a minor suggestion... it would be easier to understand when you highlight those letters when saying it my regards

[@muazzamalikhan9401]

CA=CD+AD 6=CD+AG+DF 5=CE+BH+EF ADDING ABOVE EQUATIONS 11=CD+CE+3+DE CD+CE+DE=8 THIS IS DONE BY A HIGH SCHOOL STUDENT. TRIED TO BE SIMPLE THANKS!!🙏

[@ManishKumar-sw8yk]

In other way, we know AD+BE= DE+ AB... In this way we proceed further and get the answer as 8. Your method full explanation to the problem. But in competitive exam we have less time. Thank u 4 ur work.

[@twistedsim]

Yay longer non viral video :D, thanks !!

[@tomgreg2008]

Right on! I'm getting better at these geometry ones!

[@kushagratripathi113]

Excellent channel sir Your explanations are out of the world, keep doing this.

[@poonamkulung7042]

Why did he write p/2 at 5:51? Do you know?

[@kushagratripathi113]

@@poonamkulung7042 Here perimeter= 2CF So, CF = P/2 here p- perimeter.

[@matejriha1108]

That was the hardest matematic problem ive ever seen! I had to play it three times to get it.

[@crazym108]

I gave every little segment a variable. I needed 7 variables, but then I just started spamming equations of sums of each segment and substituting. Eventually the answer of p=8 just dropped out of the sums in two different ways!

[@Nithesh2002]

I’m glad I learnt something new, i.e. showing an approach I didn’t see. However I feel you could have summarised the solution in about 10 seconds... maybe you could work on keeping simplicity.

[@abhijeetpal9858]

Hi Sir It's me Abhijeet from India. And this time I really figure it out this problem. I mean I try to solve your picked problems and I did it. Continue your amazing work. I really love your KZbin Channel Mind your Decisions.

[@TuberTugger]

BH = 5 - p/2 and AH = 6 - p/2. Also BH + AH = 3 11 - p = 3 Definitely didn't get this one, but I like this math for the ending better. Makes more sense to me.

[@timguy2463]

It's an easy problem to solve. If you use concept of tangents subtended by a point having equal lengths, it's obvious you only have to subtract tangents subtended by points A and B. Tangents subtended by other points need not be considered as they add and subtract each other. I must say maths is about finding correct, efficient solution albeit a simple one. There's no harm if solution is simple but I have noticed in your videos at times that you project problems as being more complicated than they truly are. Nothing is more beautiful than a simple solution in maths.

[@pbierre]

I solved this using a geometric computation system with graphics. I solved the magnitude of the triangle's smallest angle using Law of Cosines (theta = 0.5223148 radians). The origin was put at this vertex, v1. I generated direction vectors dirVec(theta/2) and dirVec(-theta/2), then scalar-multiplied these directions by 6 and 5 to pin down v2 and v3: v1 = [ 0 , 0 ] v2 = 6 * dirVec(theta/2) v3 = 5 * dirVec(-theta/2) I formed triangle(v1, v2, v3) and generated the incircle by algorithm. I constructed the tangent line as a vertical, at negative distance cir.r from cir.c, intersected this line to get the endpoints of the line segment, then added up the perimeter. Result: 8.0000000

[@spafon7799]

Alternatively. Define point I as the tangent of DE to the circle. Define a=AF=AG b=BG=BH c=CH=CF. We get a+c=6 b+c=5 and a+b=3. Solving these 3 eqns/3 unknowns we get a=2 b=1 and c=4. Now define d=DF=DI and e=EH=EI. We know that CF=4, and c=CF so c=d+f=4. We know that CH=4 and c=CH so c=g+e=4. Adding the last two results gives d+e+g+f=8. And d+e+g+f is the perimeter of the triangle CDE, so it equals 8.

[@chocolatechip12]

I created the tangent points and gave each unknown length a variable. That gave me the equations a+b=3, b+c+e=5, and a+d+f=6. That means a+b+c+d+e+f=11, and since I need to solve for c+d+e+f, just subtract 3 for a+b.

[@sandeepbattina2248]

Add more videos ,ur expanation is good.thanku for bringing questions of good math thinkers

[@rickgraham8701]

Deduced CF then used trig identities to get 4, * 2 = 8.

[@golfchan3613]

Trig identities? Don't they all(or at least most, because I only memorized sin/cos/tan(tangent rule or riot kappa) 30/45/60 and 90 and 0 should be obvious) need calculators? He states these people weren't allowed to use calcs, but I guess since we're doing it at home you can do that

[@mishu3245]

When I see theese problems, my love for marhs is increasing. Thanks!

[@zuregayeta7183]

How did you guys get the measurement of DE? I need an answer ASAP.

[@justpaulo]

DEC=8 I used the property that the distances from a triangle vertex to the two nearest Incircle touchpoints are equal. Let's call the 3 touchpoints F, H & I for the upper, left and lower points respectively. Also notice that the perimeter of DEC, from the problem, doesn't depends on the tangent angle, therefore, in the limit, DEC = 2*FC = 2*IC. Using the property above we proceed to find FC (or IC) like this: 1) FC = AC - AF = 6 - AF 2) AF = AH = AB - HB = 3 - HB => FC = 6 - ( 3 - HB ) 3) HB = BI = BC - IC = 5 - IC = 5 - FC => *FC = 6 - [ 3 - ( 5 - FC ) ] => FC = 4* Therefore DEC = 8.

[@herlanggaizul6965]

What's the integration of sqrt tan x dx already you explain ? I wondering about that

[@HeroicJay]

Once you take the tangent rule into account, you don't even need to double CF or introduce p or any of those other complications you add. Just recognize that - since DE=DF + DG, and AB=AF + BG, and the obvious "length of a line segment = lengths of partial line segments added together" - the perimeter of CDE is simply AC + CB - AB = 6 + 5 - 3 = 8.

[@botfeeder]

A simple way to do it is to note visually that if you start with lines AC and BC and flip the left and right segments (from the tangent point) of AD and BE using the tangent segments rule, then you wind up with line segment AB and triangle DEC. In other words, an equation for line lengths is AC+BC = AB + DEC. We want the length of triangle DEC: DEC= AC+BC-AB = 6+5-3=8, which is the answer.

[@JLvatron]

Just brilliant!

[@riffraff11235]

I did it in a more complicated way. I figured out that the area of triangle ABC is equal to 1/2 * (radius of the inscribed circle) * (perimeter of ABC). Once I got the area from Heron's formula, I solved for the radius of the circle. Knowing the area of ABC, I divided it by 3 to get the height of ABC above side AB. Assuming that the perimeter of CDE doesn't depend on the point where DE is tangent to the circle, I took DE to be parallel to AB, making ABC and DEC similar triangles. This makes it so that the points of tangency of AB and DE are connected by a diameter of the circle, so subtracting twice the radius of the circle from the height of ABC yields the height of DEC. Since the ratio of the heights of two similar triangles is equal to the ratio of their perimeters, I took the ratio of the heights and multiplied it by the perimeter of ABC to get the perimeter of DEC.

[@kingcrimson1631]

hhhhh, I managed to come up with 8 but I took a much more complex and completely different path to the answer. Great video and interesting to see the most elegant solution!

[@chellurivenkatasatyanaraya240]

Sir,in this video is very useful for all learners about mathematics, thanking you sir:-CHVSN as a INDIAN mathematician but till I am a learner of fundamental mathematics

[@Nomnomlick]

Split AB into x and 3-x at the tangence point. Now splite AC in x and 6-x at the tangence point and AB into 3-x and 5-(3-x). Solve for 6-x=5-(3-x) to find x=2 which makes the distance from C to the tangence point 4. Solved the rest by using an extreme case. The problem suggests that no matter the point of tangence of DE, the perimeter stays the same. So I chose to let the point of tangence be on BC, making D exactly C and E on the point of tangence. So the 'triangle' CDE has 2 sides of 4 and 1 side of 0 making its perimeter 8.

[@at9456]

Its actually way easier once you realize about the tangents distances, by that rule once figured out that the perimeter is CF + CG you can also realize that AF + BG equal to AB, so you can do CB + AC - AB = 6 + 5 - 3 = 8

[@jessstuart7495]

Another way, is to divide up the shape along the tangent points, and just write out the equations for the lengths of each side. (call point K the tangent point on the length 3 side) simplify the variable names... AK = AF = a BK = BG = b DF=DJ = d EG = EJ = e a + d + DC = 6 b + e + EC = 5 a + b = 3 add the first 2 equations and subtract the third.. a + d + DC + b + e + EC - a - b = 6 + 5 -3 d+e + DC + EC = 8 = Perimeter of the small triangle.

[@52gt]

For once I solved one of these problems almost immediately. When it was clear any tangent line DE yielded the same perimeter, I realized the limit of such a triangle was 2 times the line from C to the intersection of the circle and that line. The rest was algebra.

[@victorkokta]

very nice!

[@Mike80097]

I’m wondering about a more basic problem namely given the same information calculate the radius of the inscribed circle. I enjoy your channel Presh. Thank you

[@yurenchu]

Let's give short variable names to the known side lengths of the triangle: a = |BC| = 5, b = |CA| = 6, c = |AB| = 3. If you draw line segments from the center of the circle to the three vertices of the triangle, you'll divide the triangle into three smaller triangles. Each smaller triangle has a known base length (a, b or c) and an altitude that is equal to r (= the radius of the circle; because the circle is touching the sides of the triangle and hence the radii are perpendicular the sides; so they are indeed altitudes of these smalller triangles). So the area of triangle ABC equals the sum of areas of these three smaller triangles: Area = r*a/2 + r*b/2 + r*c/2 = r*(a+b+c)/2. However, the area of triangle ABC can also be calculated by Heron's Formula: Area = sqrt( s*(s-a)*(s-b)*(s-c) ) in which s = (a+b+c)/2 . So we now have two formulas for calcultating the area of ABC, we can equate them and from there solve for variable r. r*(a+b+c)/2 = sqrt( s*(s-a)*(s-b)*(s-c) ) ... substitute (a+b+c)/2 = s in lefthandside ... r*s = sqrt( s*(s-a)*(s-b)*(s-c) ) ... square both sides .... r² * s² = s*(s-a)*(s-b)*(s-c) r² = [ (s-a)*(s-b)*(s-c) ]/s So in this particular example, with a = 5, b = 6, c = 3, we have s = (5+6+3)/2 = 14/2 = 7 r² = [ (7-5)*(7-6)*(7-3) ]/7 r² = 2*1*4/7 = 8/7 r = sqrt(8/7) = (2/7)*sqrt(14) There you go!

[@akameel]

Excellent 👍🏽

[@jbtechcon7434]

"Hey, this is Presh... Talwalkar? And you're watching MindYourDecisions..."

[@chessandmathguy]

JBTechCon lol I didn't notice that at first.

[@jbtechcon7434]

I didn't notice it at first either, now I can't unhear it. :-)

[@reddcube]

WOW. clever solution. Personal I would can length CF as 'h' instead of 'p/2'. moving all the halves around means I'm more likely to mess up.

[@VarunSingh-lc7oz]

Sorry for the spelling mistake of EASILY * But I solved it by tangent theorem

[@Moradabadi_inspector]

Cgl mains me aate h ye questions

[@harshilpanchal2229]

How did you do by tangent theorem. I mean what did you do with the right angles

[@mathswithmanojkumargathain_134]

Sir...... salute you 👌👌👌👌👌👌

[@geoffreytrang8670]

In fact, if you draw any two tangent lines from a common exterior point and the corresponding radii, you will always get a kite. The kite will always divide into two congruent right triangles via one of the diagonals, and have a circumcircle. A general kite need not have a circumcircle (e.g. a rhombus that is not a square), but will always have an incircle.

[@roderickwhitehead]

Excellent problem.

[@floyo]

I used the second way, but I prefer your first solution, because it also proves that the triangle keeps the same perimeter.

[@GermansEagle]

just use the fact that the vertices create equal sides that tangent to the circle(lets say, you put AB as a + 3-a, it solves itself) . It you will get AC is 2,x and 4-x, and BC is 1,y and 4-y. x and y also make the tangent between AC and BC, so it is x+y. So then it is 4-y + 4-x + (x+y) = 8 Easy as that

[@drskmobinulhaque7554]

Some of your geometry problems are from the book 'Challenging Problems in Geometry' by Alfred Posamentier. This problem is one of them. I have that book.But the problem in the book involves proving it. This is the first problem of the fourth chapter (Circles Revisited).I like your channel.Keep uploading new videos.

[@yvessioui2716]

That is a clear explanation. I like it. While tinking at this problem I was wondering what is the relation between are, thus radius, of a circle inscribed in a specified triangle (known by its 3 lenghts) on a general approach? Something worth for any triangle in fact. I look for it and found on google that area of circle = 1/2 r times triangle perimeter. This answers that. And then I went to 'Why don't use this opportunity to find the same kind of small triangle perimeter in each corner of the big triangle and compare all those areas added with the circle or with area of the big circle in general terms?' I will do that in my spare time if I ever find some. I hypothesize there is kind of a constant there for any given big triangle since there can be only one inscribed circle in a specified triangle.

[@bbgg1690]

This was easy but thanks for your hard work bro.

[@drmichaelelinski6992]

It’s been so long since I’ve taken geometry, that forgotten the basic theories and formulas for geometric figures. Where can I brush up on these geometric assumptions?

[@sameerkumar3865]

That's awesome man

[@geoninja8971]

This might be the first MYD video where I think my solution is more elegant... I often get there in the end in a mess of decimals and trig, but this one just appeared in front of me....

[@davidwillis7991]

same

[@zahidluqman8202]

By similiarity of triangles or by pythagoras in the triangle in the circle we gwt cd=cg Ac-Af=Bc-Bg Ac=6,Bc=5. Af=Bg+1 AB=Af+Bg.subst We get Af=2,Bg=1 Peri=cd+ce+de 4-y, 4-x ,x+y.subst. =8

[@suvankardasgupta2074]

Unbelievable... I am inspired...

[@ianmangion4760]

Hey Presh, I got to the answer faster by the below reasoning: p=CE+CD+DE as DJ=DF and EJ =EG therefore: p=CF+CG p=(AC-AF)+(BC-BG) p=AC+BC-(AF+BG) since AF=AH and BG=BH and BH+AH=AB we can also say that AF+BG = AB thus: the equation is simply: p=AC+BC-AB p=6+5-3 p=8 what do you think about this reasoning?

[@Mortagus]

the background colors in the miniature is awesome, you should use the same color theme in the video ^^

[@ml8727]

Before seeing the answer. AB=a+b=3, BC=b+c=5, CA=c+a=6, find c. c=4. The perimeter of CED is 2c, or 8.

[@billrussell3955]

That was great!

[@zralok]

Something to say bout the Känguru-contest is you have three different types of difficulty for problems A,B,C and for each there are 10 problems. This one was a problem of type C. And another thing to know is that it is multiple-choice. So you have 5 different answers where you then write the letter for the correct solution on your answer ticket. (I hope my english is not as bad as I think it is xD)

[@mkb6418]

It was farely easy, if you just made the drawing. Constructing a proper drawing is underestimated most of the times, but it's a really important step.

[@mysticdragonex815]

Finally a question here that I could solve

[@brunoa.colturato5868]

I got this one. Great!

[@rec1966]

I solved it differently by working out that CDE’s perimeter is equal to ABC’s perimeter minus 2AB because the 2 tangents at A are equal, the same with B, which adds to 2AB, then the tangents at D and E are equal and cancel each other out.

[@bpark10001]

Once you use tangent theory to establish that the triangle's perimeter is equal to length FCG, you need only take the perimeter of the entire large triangle (5+6+3) and using the tangent theory again, see that 3 gets subtracted out twice from that (5+6+3 -(3+3)). You don't need to know that FC = CG.

[@1ClassicalMusicFan]

For viewers who are not aware of the "Pitot Theorem": (I) . At 5:46, the answer is 2(CF). ***** Now, "to figure out CF" has nothing to do with DE. (II) Back at 3:10, without even considering DE (assuming that D, E and "tangent segment DE" were not there), based on what has been covered up to 3:10, it is easy to get CF = 4. (A lot of viewers must have done (II) before.)

[@yurenchu]

"(A lot of viewers must have done (II) before.)" Yes, and then I... euh, I mean "they", got stuck. :-D

[@nightstar8444]

Is the cd=cf is miraculously happen, or is it always happen to any kind of this line?

[@txaxneverdie145]

I feel like I'm missing something, how is it AF=AH and its p/2 - 2. But it's not 6 - p/2 ? Where did the - 2 come from?

[@Tehom1]

Thanks for the challenge, Presh! To solve it, let's first use the fact that from a point outside a circle, the two tangents that you can draw have the same length from the point to the circle. With that in hand, we can subtract and add pairs of identical distances to get the answer. Let's label the point on the circle, between D and A and on the line DA as "da", and similarly for points "de", "ab", and "be". Then by partitioning line segments we have: CD = CA - Dda - Ada CE = CB - Ebe - Bbe DE = Dde + Ede AB = Aab + Bba And by the tangent equivalences, we have: Dda = Dde Ede = Ebe Bbe = Bba Ada = Aab So with substitution, we have: CD = 6 - Dda - Aab CE = 5 - Ebe - Bba DE = Dda + Ebe 3 = Aab + Bba Giving us: CD + DE + CE = 6 - Dda - Aab + Dda + Ebe + 5 - Ebe - Bba = 11 - Aab - Bba = 11 - 3 = 8 Good puzzle, Presh!

[@PeterNjeim]

You used an overly complicated solution. Another way to solve this, which is much easier and faster, is to use the Pitot theorem. Here is an explanation by Adam Stanisławski: _"The problem becomes trivial when you use Pitot theorem ("in a tangential quadrilateral (i.e. one in which a circle can be inscribed) the two sums of lengths of opposite sides are the same" - from Wikipedia). That means that AB + DE = BE + AD and therefore DE = BE + AD - AB. The perimeter of desired triangle is EC + DC + ED = EC + DC + (BE + AD - AB). Since BE + EC = BC = 5 and AD + DC = AC = 6, perimeter equals BC + AC - AB = 5 + 6 - 3 = 8."_

[@rmsgrey]

Complexity of proof is a tricky subject - using established theorems hides complexity, but doesn't remove it. So solving via Pitot means you're implicitly including a proof of Pitot in your solution process... If there's a theorem that says that the perimeter of the triangle formed by a tangent to a triangle's incircle and the obvious vertex is equal to the length of the two sides of the original triangle meeting at that vertex less the third side, then invoking that theorem would make the problem entirely trivial, but wouldn't make the solution any less complicated - just hiding the complexity behind the theorem...

[@Tehom1]

In any case, comments on this puzzle have introduced me to Pitot's theorem, so that's good.

[@Tehom1]

Reminded of somebody's humorous proof formulation, "Assuming the entire known body of mathematics, it trivially follows that ..."

[@shubhangdadhich9908]

Pivot's theorem reduces effort to 20seconds max. And it's taught in school too

[@davidwillis7991]

Not knowing that theorem meant it took me 30 seconds

[@vqlim]

Hi Presh, in this solution, if you draw the 90 degree line from center of the circle to the 3 lines of triangle, you can solve it in other concept 😁

[@MrBrain4]

I feel I have a much simpler solution: Since we know the perimeter of the triangle is equal to CD+CG, and since AD+BG=3, then the perimeter is simply 5+6-3, which is (AC+BC)-(AD-BG).

[@kidwithaphonecamera]

There is another method. So in ADBE sum of opposite sides is constant (One of theorems in geometry, can be proved using equal tangents) AB+DE=BE+AD. Hence: DE=BE+AD-3. But: CD=6-AD. And: CE=5-BA. Adding this up: 6+5-3=8.

[@saxbend]

Do a Papa Flammy spoof. You know you want to.

[@Abdega]

Yes *MATH BEEF!*

[@PranavMurukesh]

Wonderful problem