Mock AIME 3 Pre 2005 Problems/Problem 9

Problem

$ABC$ is an isosceles triangle with base $\overline{AB}$ . $D$ is a point on $\overline{AC}$ and $E$ is the point on the extension of $\overline{BD}$ past $D$ such that $\angle{BAE}$ is right. If $BD = 15, DE = 2,$ and $BC = 16$ , then $CD$ can be expressed as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Determine $m + n$ .

Solution 1

Let the foot of the altitude from $C$ to $AB$ be point $F$ and from $D$ to $AB$ be point $G.$ $\triangle BAC \sim \triangle BAE.$ Since $\triangle ABC$ is isosceles, $\frac{BE}{AB} = \frac{1}{2},$ so $BX = \frac{17}{2}.$ Then $DX = 17 - 2 - \frac{17}{2} = \frac{13}{2}.$ Notice that $AG$ corresponds to $2$ , $GF$ corresponds to $\frac{13}{2},$ and $BF$ corresponds to $BX$ by some scale factor $k$ . Also note that $\triangle AGD \sim \triangle AFC.$ Then we have $\frac{AG}{AF} = \frac{AD}{AC}$ $\implies \frac{2k}{AB - BF} = \frac{AD}{16}$ $\implies \frac{2k}{17k/2} = \frac{AD}{16}$ $\implies \frac{4}{17} = \frac{AD}{16} \implies AD = \frac{64}{17}.$ Then $CD$ is $16 - \frac{64}{17} = \frac{208}{17}.$ $208 + 17 = \boxed{225}.$

~grogg007

Solution 2 (fakesolve, don't do this)

Let the perpendicular from $C$ intersect $AB$ at $H.$ Let $CH$ intersect $BD$ at $P.$ Then let $AP$ intersect $BC$ at $F.$

Note that $\triangle AEB\sim \triangle HPB,$ with a factor of $2.$ So $BP=8.5$ and $DP=6.5.$ Then the mass of $P$ is $15$ and the mass of $D$ is $8.5$ and the mass of $B$ is $6.5.$ Because the triangle is isosceles, the mass of $A$ is also $6.5.$ So $CD=\frac{8.5}{8.5+6.5}\cdot 16.$

Mock AIME 3 Pre 2005 (Problems, Source)
Preceded by Problem 8	Followed by Problem 10
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Mock AIME 3 Pre 2005 Problems/Problem 9

Contents

Problem

Solution 1

Solution 2 (fakesolve, don't do this)

See Also