Area Of An Innocent Right Triangle

Geometry Level 3

A right triangle has a hypotenuse of length 20 and the length of the altitude dropped to the hypotenuse is 16. What is this triangle's area?

20 100 160 320 This question is flawed.

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7 solutions

Rahul Saha
Jan 10, 2016

It can be shown using AM-GM that the altitude to the hypotenuse is at most half of the hypotenuse. It follows that there is no such triangle, and hence the correct answer is This question is flawed \boxed{\text{This question is flawed}} .

Source: A post on MSE

To quote @Calvin Lin,

...This is one reason why I stress considering the possibility of the geometrical setup and ensuring that it actually exists.

The main point of checking the existence of the possibility of a geometric structure isn't necessarily that the question is deliberately trying to fool you (like this one), but that checking why a geometric structure exists will give you a pointer towards some narrow but definite parts of a diagram that have more restrictions imposed on them (otherwise the diagram would cease to exist). These parts of a diagram should theoretically be easier to investigate due to having more restrictions on them. This is still true in olympiad maths, not just pre-olympiads.

Kaustubh Miglani
Jan 12, 2016

If the area of triangle is 160.Then it means if a,b are legs of the triangle then a b 2 = 160 \frac{ab}{2}=160 it follows that ab=320 and 2ab=640 and a 2 + b 2 a^2+b^2 =400 (Pythagoras theorem).Subtracting 1st equation from 2nd we get a 2 + b 2 2 a b a^2+b^2-2ab =-240 as a 2 + b 2 2 a b a^2+b^2-2ab is a square of (a-b) and square of anything cant be negative.It isnt possible.

Why did you assume in the beginning that the area of the triangle is 160?

Nate Gillman - 5 years, 5 months ago

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i didnt assume area of triangle=b*h/2 use that to get area=320/2=160

Kaustubh Miglani - 5 years, 5 months ago

From the diagram above, we get:

A B = 1 6 2 + x 2 AB = \sqrt{16^{2}+x^{2}} and B C = 1 6 2 + ( 20 x ) 2 BC = \sqrt{16^{2}+(20-x)^{2}} .

Then, combining them and simplifying, we have x 2 20 x + 206 = 0 x^{2}-20x+206=0 , which has no real solutions.

Wow!This is better solution :)

arifuzzaman arif - 5 years, 5 months ago

Combining means

Suneel Kumar - 5 years, 5 months ago

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Note that by Pythagoras Theorem, A B 2 + B C 2 = 400 AB^{2}+BC^{2}=400 .

A Former Brilliant Member - 5 years, 5 months ago
Ali Qureshi
Jan 16, 2016

Consider the right angled triangle with its hypotenuse as the diameter of a circle. Clearly the altitude cannot be more than the radius of circle. The radius of circle is half of hypotenuse 10.

In my answer, I note that the proof can be given by AM-GM. Your answer is actually a "proof without words" of the two variable case of AM-GM.

Rahul Saha - 5 years, 5 months ago
Henny Lim
Jan 10, 2016

The possible setup for right triangle with hypotenuse of 20: 12 and 16. If the hypotenuse was reduced into 16, then the setup would be impossible as there are two sides with 16 of length. Therefore the triangle should not be right anymore, but it will become isosceles one.

Circumscribe the triangle into a circle or radius 20 and acording to the figure x + y = 20 x+y = 20 by M A M G MA-MG we see that 20 = x + y 2 x y 10 x y 20=x+y \geq 2\sqrt{xy} \Rightarrow 10 \geq \sqrt{xy} Therefore there is not a triangle with this measures.

Paulo Filho
Jan 13, 2016

In any right triangle, the altitude divides the hypotenuse in two parts, with respective lengths x and y, satisfying:

x + y = length of hypotenuse

x * y = (length of altitude)²

In this problem:

x + y = 20

x * y = 16² = 256

This system of equations has no real solution (the graphs of each equation don't touch each other).

Therefore, this right triangle is not possible.

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