Can You Solve This MIT Admissions Question? Geometry Problem, 1869

Can You Solve This MIT Admissions Question? Geometry Problem, 1869

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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19 thoughts on “Can You Solve This MIT Admissions Question? Geometry Problem, 1869

  1. I reached the point where p/16 = 9/p with Trigonometry but then I didn't know what to do after. Great Video!

  2. Not hard…. I'm guessing this was one of the easier questions. Any smart 12-14 year old should get the answer.

  3. It can be solved too if you just know that in a rectangle triangle the vertex of the right angle lies in a circle whose center is in the middle point of the oter two vertex. In that case:


    The ecuation of a circle with r=12.5 is


    In the Point of Interest x= 16-12.5 = 3.5

    We can obtain y (p in the point of interest) by

    y = sqrt(12.5^2-3.5^2) = 12

    I cant guess why physicists like me are supposed to be weird people… ¯_(ツ)_/¯

  4. based on the catheus theorem the altitude of a right triangle towards the hypotenuse equals the square root of the the product of the 2 lengths of the hypotenuse split by the altitude

  5. Since the big triangle is a right triangle we could use the formula p = 1/2*25= 12.5
    Once we find p, the rest is easy

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