A geometry problem by Matheus Yuzawa

Geometry Level 4

Consider a right triangle ABC with a 90° angle in A. Be AE and AD the height and the median relative to the hypotenuse BC, respectively. If the median BE measure is $(\sqrt{2} - 1)$ cm and AD measure is 1cm, what's the measure of AC?

(This problem was taken from ITA-Brazil 2013-2014 entrance exam)

This question is flagged because it is unclear what point $E$ is meant to be.

3 - \sqrt{2}

\sqrt{6 - 2\sqrt{2}}

4\sqrt{2} - 5

3\sqrt{4\sqrt{2} - 5}

1 solution

Aaaaaa Bbbbbb
Aug 2, 2014

$BE=\sqrt{2}-1$ AD=1, BC=2 $CE=2-(\sqrt{2}-1)=3-\sqrt{2}$ $AB^2+AC^2=4$ $AC^2-AB^2=CE^2-BE^2=(3-\sqrt{2})^2-(\sqrt{2}-1)^2$ $=11-6\sqrt{2}-3+2\sqrt{2}=8-4\sqrt{2}$ $2AC^2=12-4\sqrt{2}$ $AC=\boxed{\sqrt{6-2\sqrt{2}}}$

@Aaaaaa Bbbbb how is BC=2 ?????

Rishi Hazra - 6 years, 10 months ago

Because $AD=\frac{BC}{2}$

aaaaa bbbbb - 6 years, 10 months ago

will you please elaborate?

Rishi Hazra - 6 years, 10 months ago

We know that the hypotenuse is the longest leg in a right triangle, but how come the leg is longer than the hypotenuse? I found the values of each of the choices and only one option is smaller than the hypotenuse value, so i thought that it was the answer, but how come?? Will you please explain your answer? In an understandable manner please..

Mark Vincent Mamigo - 6 years, 7 months ago

@Matheus Yuzawa The clarification request stated "If AE is the height then how could BE be the median?"

Please clarify which is point E, thanks.

Calvin Lin Staff - 6 years, 10 months ago

Hi, I was checking the original question and there's a mistake in the text. In the original question it says: "If BE measure is" instead of "If the median BE measure is", that may solve the problem. I am Matheus' friend and do the translations of these questions with him, sometimes things like this happens, sorry. Can you edit it? He is quite busy currently.

Nick Okita - 6 years, 6 months ago

A geometry problem by Matheus Yuzawa

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