The Massachusetts Institute of Technology (MIT) is one of the top ranked universities in the world. This question appeared on its admissions exam nearly 150 years ago. “The perpendicular dropped from the vertex of the right angle upon the hypotenuse divides it into two segments of 9 and 16 feet respectively. Find the lengths of the perpendicular, and the two legs of the triangle.” The video presents a solution.

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Sources

MIT Questions: https://libraries.mit.edu/archives/exhibits/exam/geometry.html

MIT Answers: https://libraries.mit.edu/archives/exhibits/exam/geometry-answers.html

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DOES ANYONE HERE EVER HEARD ABOUT EUCLID'S THEOREM?

Wow, that's way simpler – I made the hypotenuse the diameter of a circle, and had to fiddle around with fractions.

This was really easy to solve coming from an eight grader. To find p, I just took the geometric mean of 9 and 16.

sqrt(9*16)

surprisingly this is a much easier question that I would have anticipated to be used in MIT's entrance exam… I am not a mathematician and yet I can think of sbout 4 different ways to solve this!…

A much much much easier way to do this is by using the Geometric mean. It only takes a few seconds and you don't have to worry about mismatching the corresponding sides of the similar triangles

I couldn't solve this, so I called my Russian friend Grigori Perelman, and told him I'm going to give him 1 million dollars if he solves this puzzle for me… He did it for free. :)

forgot about symmetrical triangle theory.

i've solved it by using

tan∆=9/x and

tan(90-∆)=16/x

ie. cot∆=16/x

next, 9/x= x/16

x=12, where x is the vertex.????

It's very easy. Left side is 15 and bottom 20, solve it in 1 minute

Greeks solved this 2500 years ago

Maybe it's not the simplest solution, but we can count that with tree euqation of pythagorean theorem:

9^2 + p^2 = a^2

16^2 + p^2 = b^2

a^2 + b^2 = 25^2

a^2 + 16^2 + p^2 = 25^2

9^2 + p^2 + 16^2 + p^2 = 25^2

81 + 256 + 2p^2 = 625

2p^2 = 288

p^2 = 144

p = 12

a^2 = 81 +144 => a = 15

b^2 = 265 + 144 => b = 20

i forgot similar triangles, although i noticed 3,4,5 ratio before i even began to solve.

so what?

I learned this at school when I was 13.

This is just Pythagoras theorem 3 times then solve the easy equations.

This is waaay to easy for an admissions test, even if it was from 1869

I learned this math in 8th grade

All you need is use Pythagoras theorem two times. Is not that hard to be fare.

you can also do it with base grade pythagoras

25^2=a^2+b^2. a=16 b=20. 9^2+p^2=16^2. p=12.

No fancier math needed. Remember, this was written for M.I.T. in 1869. See problem, see solution in flash.

i am a junior high school student .And I solved it……

Fun Q. I like tricky questions with easy solution.

I did it by creating 3 equations

X^2 = P^2 + 9^2

Y^2 = P^2 + 16^2

25^2 = X^2 + Y^2

easy , use the 3-4-5 rule dont need any calculator

Very easy

This was wicked easy. Like a simple geometry question

You can solve the legs with an easier way. The right leg squared is 16*(16+9) and the left leg squared is 9*(9+16). You can use ratios too but i find this way easier.

So in 1869 you could be admitted into MIT with the knowledge you get in 9th grade? Wow

Those MIT questions seems to be too easy, actually, have you ever tried to solve the questions of the brazilian admissions tests for militaries academies? Such as IME and ITA

You can solve this question by using the pythagorean theorem as well. It took me a little longer but I got the same results.

I didn't realize the 345 pattern when I did it but it was still fun.

Thanks for the video!

you can get the hypotenuse by geometric mean method I. e the square of hypotenuse = the sides divided by the hypotenuse are multiplied. and you will get the sides of right angled triangle by trigonometry considering the angle theta

I got it.

Thank you for your blog. I was a "successful" Algebra-1 teacher at S.M. White Middle School, Carson, CA in 2005-06 because my Uncle Larry Holland, who graduated from MIT, told me how he earned a high GPA for his BA Mathematics degree without physically writing out one mathematics problem. He did this amazing feat at an amazing university by thoroughly explaining verbally and in written form how to solve every math problem in every test he took at MIT. When you can explain how to do a math problem you don't need to actually do it because it becomes an unnecessary trivial activity. To qualify to teach mathematics in California every Mathematics teacher have to do the same thing as my Uncle Larry did but only to explain each component of the Quadratic Formula. This formula only exists to create a reason to learn most of the basic steps to correctly calculate the answer of a quadratic formula given the least amount of information necessary to find its answer.

pretty easy tho..haha

fuck!!! such a easy question!! in India they teach us this in 9 th grade!!!!

I'm 14 and I solved this with only phytagoras, its really easy

its a 3,4,5 triangle

p(2)=9*16=144 p=12 very easy !!!In Vietnam children can solve this before they go to high school (grade 7)

I recall that my old math teacher did this once. But it is great that you refresh my memory .

U COULD SOLVE THIS PRBLM IN 60 SEC…USING PYTHOGORAS THEORAM THAT IS 3:4:5..SIMPLE

All these comments about it being so easy are from people that live in shithole countries where they still have malaria and raw sewage running in their streets. Have fun knowing you could have gotten into MIT in 1869 as you sew Nikes together and shit in a hole in the ground.

Your method is certainly efficient when one recognizes the triangles are similar. I should have but I did not. This problem can also be solved by writing the Pythagorean theorem for all three triangles. The two equations for the two smaller triangles can be used to eliminate the variable P. This equation and the equation for the larger triangle can be used to eliminate either variable representing a leg of the larger triangle. This leaves one equation in one variable. By substituting the numerical value determined for one leg into the equation for the larger triangle the value of the other leg can be determined. With either leg of the larger triangle known the value of P can be determined. Obvious this is a brute force method but it achieves the same results.

i solved it