Mock AIME 1 2007-2008 Problems/Problem 11

Problem

$\triangle DEF$ is inscribed inside $\triangle ABC$ such that $D,E,F$ lie on $BC, AC, AB$ , respectively. The circumcircles of $\triangle DEC, \triangle BFD, \triangle AFE$ have centers $O_1,O_2,O_3$ , respectively. Also, $AB = 23, BC = 25, AC=24$ , and $\stackrel{\frown}{BF} = \stackrel{\frown}{EC},\ \stackrel{\frown}{AF} = \stackrel{\frown}{CD},\ \stackrel{\frown}{AE} = \stackrel{\frown}{BD}$ . The length of $BD$ can be written in the form $\frac mn$ , where $m$ and $n$ are relatively prime integers. Find $m+n$ .

Solution

$[asy] size(150); defaultpen(linewidth(0.8)); import markers; pair B = (0,0), C = (25,0), A = (578/50,19.8838); draw(A--B--C--cycle); label("$B$",B,SW); label("$C$",C,SE); label("$A$",A,N); pair D = (13,0), E = (11*A + 13*C)/24, F = (12*B + 11*A)/23; draw(D--E--F--cycle); label("$D$",D,dir(-90)); label("$E$",E,dir(0)); label("$F$",F,dir(180)); draw(A--E,StickIntervalMarker(1,3,size=6));draw(B--D,StickIntervalMarker(1,3,size=6)); draw(F--B,StickIntervalMarker(1,2,size=6)); draw(E--C,StickIntervalMarker(1,2,size=6)); draw(A--F,StickIntervalMarker(1,1,size=6)); draw(C--D,StickIntervalMarker(1,1,size=6)); label("24",A--C,5*dir(0)); label("25",B--C,5*dir(-90)); label("23",B--A,5*dir(180)); [/asy]$

From adjacent sides, the following relationships can be derived:

$\begin{align*} DC &= EC + 1\\ AE &= AF + 1\\ BD &= BF + 2 \end{align*}$

Since $BF = EC$ , and $DC = BF + 1$ , $BD = DC + 1$ . Thus, $BC = BD + DC = BD + (BD - 1)$ . $26 = 2BD$ . Thus, $BD = 13/1$ . Thus, the answer is $\boxed{014}$ .

Mock AIME 1 2007-2008 (Problems, Source)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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Mock AIME 1 2007-2008 Problems/Problem 11

Problem

Solution

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