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July 27th, 2015, 04:35 AM   #1
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Maximum area of a square inscribed in a triangle

I'm really stumped on this problem.

I have worked out the three smaller triangles inside of the bigger triangle but after that, nothing else at all.
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July 27th, 2015, 05:37 AM   #2
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Let $\displaystyle a$ = horizontal length of base of lower-left triangle
Let $\displaystyle b$ = horizontal length of base of lower-right triangle
Let $\displaystyle \theta$ = angle at lower-right of diagram (large triangle)

Therefore, $\displaystyle a+b = 60$

Looking at lower-right triangle, we have

$\displaystyle \sin\theta = \frac{x}{b}$
$\displaystyle b = \frac{x}{\sin\theta}$

Looking at the lower-left triangle, we have

$\displaystyle \cos\theta = \frac{a}{y}$
$\displaystyle a = y \cos\theta$

Therefore:

$\displaystyle a+b = y \cos\theta + \frac{x}{\sin\theta} = 60$
$\displaystyle x = 60 \sin\theta - y \sin\theta \cos\theta$

Let $\displaystyle A$ = area of the rectangle:

$\displaystyle A = xy = 60y \sin\theta - y^2 \sin\theta \cos\theta$

We now have the area of the rectangle in terms of a single variable, $\displaystyle y$. Can you finish?
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July 27th, 2015, 05:53 AM   #3
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I have finished it and indeed, it is correct (we were given the answer).

Thank you so much.

I have absolutely no idea how you guys could manage to find such elegant solutions in these kinds of problems. I just get stumped.

By the way, is there an online site that provides problems like those?
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July 27th, 2015, 11:49 AM   #4
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Were you given that the original triangle is right-angled?
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July 27th, 2015, 03:49 PM   #5
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Yes, it was given that it is a right triangle.
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