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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 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,161 Thanks: 734 Math Focus: Physics, mathematical modelling, numerical and computational solutions 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?
 July 27th, 2015, 05:53 AM #3 Newbie   Joined: Oct 2012 Posts: 29 Thanks: 0 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?
 July 27th, 2015, 11:49 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216 Were you given that the original triangle is right-angled?
 July 27th, 2015, 03:49 PM #5 Newbie   Joined: Oct 2012 Posts: 29 Thanks: 0 Yes, it was given that it is a right triangle.

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# maximum area of a square inscribed in a right triangle

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