## Friday, February 16, 2007

### Common Law. Topic: Geometry/Trigonometry. Level: AIME.

Problem: (2003 AIME1 - #12) In convex quadrilateral $ABCD$, $\angle A = \angle C$, $AB = CD = 180$, and $AD \neq BC$. The perimeter of $ABCD$ is $640$. Find $\lfloor 1000 \cos{\angle A} \rfloor$.

Solution: Well, let's say that $BC = x$ and $AD = y$. We know $x+y+180+180 = 640 \Rightarrow x+y = 280$. Consider the two triangles $ABD$ and $CBD$. Using the Law of Cosines on $\angle A$ and $\angle C$, which we know are equal (let them both equal $\theta$), we have

$BD^2 = AB^2+AD^2-2 \cdot AB \cdot AD \cos{\theta}$

$BD^2 = CB^2+CD^2-2 \cdot CB \cdot CD \cos{\theta}$.

Substituting $AB = CD = 180$ and $BC = x$, $AD = y$, it becomes

$BD^2 = 180^2+y^2-360y \cos{\theta}$

$BD^2 = x^2+180^2-360x \cos{\theta}$.

Equating the two,

$180^2+y^2-360y \cos{\theta} = x^2+180^2-360x \cos{\theta} \Rightarrow x^2-y^2 = 360(x-y)\cos{\theta}$.

Recall that $x+y = 280$, so

$x^2-y^2 = (x+y)(x-y) = 280(x-y) = 360(x-y)\cos{\theta}$.

Finally, since $x \neq y \Rightarrow x-y \neq 0$, we divide through to get

$\cos{\theta} = \frac{280}{360} = \frac{7}{9} \Rightarrow \lfloor 1000 \cos{\theta} \rfloor = 777$.

QED.

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Comment: This is a demonstration of the power of something as simple as the Law of Cosines. Never underestimate what a little experimentation can do for you on the AIME; play around with equations. If at first you do not see an approach, look at the question itself for hints. Since you have to find the cosine, it should immediately trigger the Law of Cosines because it relates sides and angles. Then apply the Law of Cosines to the important angles (the ones you know something about), in this case $\angle A$ and $\angle C$, from which the result falls quite nicely.

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Practice Problem: (2003 AIMEI - #6) The sum of the areas of all triangles whose vertices are also vertices of a $1\times 1 \times 1$ cube is $m+\sqrt{n}+\sqrt{p}$, where $m$, $n$, and $p$ are integers. Find $m+n+p$.