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In convex quadrilateral $ABCD,$ $\angle A \cong \angle C,$ $AB=CD=180,$ and $AD \ne BC.$ The perimeter of $ABCD$ is $640.$ Find $\cos A.$

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I have no idea on where to start on this problem

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Hint: Let $AD = x$ and $BC = y$. Then $x+y+180+180 = 640$ so $x+y = 280$. By the Law of Cosines, we have $$ BD^2 = AD^2+AB^2-2AD\cdot AB\cos\alpha = x^2+180^2-360x\cos\alpha $$ and $$ BD^2 = BC^2+CD^2-2BC\cdot CD\cos\alpha = y^2+180^2-360y\cos\alpha. $$ Therefore $$0 = x^2-y^2 - 360(x-y)\cos\alpha. $$ Can you take it from here?

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  • $\begingroup$ Yeah, i got 7/9. Thank you! $\endgroup$ – Human Sep 25 '16 at 2:59
  • $\begingroup$ @Human no problem! $\endgroup$ – Joey Zou Sep 25 '16 at 3:03
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Elegant Solution

Define point $E$ on the side $AD$ such that $|AE|=|BC|$. Now you have $ABE \equiv CBD$. Therefore $|BE| = |BD|.$ Now define the point $F$ as the mid point of $ED$ since $|BE| = |BD|$, we have $BF$ perpendicular to $AD$. A very nice observation here is $$ |AF| = \frac{|AD| + |AE|}{2} = \frac{|AD| + |BC|}{2} = = \frac{640 - 2(180)}{2} = 140 $$ Hence $$ \cos \alpha = \frac{|AF|}{180} = \frac{140}{180} $$

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  • $\begingroup$ In the first line, I think you meant to say "$|AE| = |BC|$", not "$|AD| = |BC|$". Nice solution! $\endgroup$ – Joey Zou Sep 25 '16 at 3:53
  • $\begingroup$ @JoeyZou Yes, thank you! Fixed. $\endgroup$ – iamvegan Sep 25 '16 at 4:03

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