Go Go Geometry! Topic: Geometry/Trigonometry. Level: AIME.

Problem: (2006 Mock AIME 1 - #15) Let $ABCD$ be a rectangle and $AB = 24$. Let $E$ be a point on $BC$ (between $B$ and $C$) such that $DE = 25$ and $\tan{\angle BDE} = 3$. Let $F$ be the foot of the perpendicular from $A$ to $BD$. Extend $AF$ to intersect $DC$ at $G$. Extend $DE$ to intersect $AG$ at $H$. Let $I$ be the foot of the perpendicular from $H$ to $DG$. The length of $IG$ is $\displaystyle \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Suppose $\alpha$ is the sum of the distinct prime factors of $m$ and $\beta$ is the sum of the distinct prime factors of $n$. Find $\alpha + \beta$.



Solution: We begin by solving for $EC = \sqrt{25^2-24^2} = 7$. Then $\tan{EDC} = \frac{7}{24}$.

By the tangent addition formula, we have

$\tan{BDC} = \tan{(BDE+EDC)} = \frac{\tan{BDE}+\tan{EDC}}{1-\tan{BDE}\tan{EDC}} = \frac{3+\frac{7}{24}}{1-3\left(\frac{7}{24}\right)} = \frac{79}{3} $.

Since $\tan{BDC} = \frac{BC}{CD}$, we have $\displaystyle BC = (CD)\tan{BDC} = 24\left(\frac{79}{3}\right) = 632$.

Notice that $\angle AGD = 90 - \angle BDC \Rightarrow \tan{AGD} = \frac{1}{\tan{BDC}} = \frac{3}{79}$.

Then $\tan{AGD} = \frac{AD}{DG} \Rightarrow DG = \frac{AD}{\tan{AGD}} = \frac{632}{\frac{3}{79}} = \frac{2^3 \cdot 79}{3}$.

Consider $\triangle HIG$. We have $\tan{HGI} = \frac{HI}{IG} \Rightarrow HI = (IG)\tan{HGI} = \frac{3}{79}(IG)$.

Notice that $\triangle EDC$ is similar to $\triangle HDI$, so $\frac{HI}{ID} = \frac{EC}{CD} = \frac{7}{24} \Rightarrow HI = \frac{7}{24}(ID) = \frac{7}{24}(DG-IG)$.

Combining our two expressions for $HI$, we get

$ HI = \frac{3}{79}(IG) = \frac{7}{24}(DG-IG)$.

Solving this for $IG$, we get

$IG = \frac{\frac{7}{24}(DG)}{\frac{3}{79}+\frac{7}{24}$.

But we calculated $DG = \frac{2^3 \cdot 79}{3}$ above, so upon substitution we get

$IG = \frac{\frac{7}{24}\left(\frac{2^3 \cdot 79}{3}\right)}{\frac{3 \cdot 24 + 7 \cdot 79}{24 \cdot 79}} = \frac{2^3 \cdot 7 \cdot 79^3}{3 \cdot 5^4} $.

Summing the distinct prime factors, we have $2+7+79+3+5 = 96$.