2006 iTest Problems/Problem 30

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Problem

Triangle $ABC$ is equilateral. Points $D$ and $E$ are the midpoints of segments $BC$ and $AC$ respectively. $F$ is the point on segment $AB$ such that $2BF=AF$. Let $P$ denote the intersection of $AD$ and $EF$, The value of $EP/PF$ can be expressed as $m/n$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

[asy] import olympiad; size(150); defaultpen(linewidth(0.7) + fontsize(11pt)); pair A = origin, B = (1,0), C = dir(60), D = (B+C)/2, E = C/2, F = 2*B/3, P = intersectionpoint(A--D,E--F); draw(A--B--C--A--D^^E--F); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,S); label("$P$",P,N); [/asy]

Solutions

Solution 1

Since $AE = EC$ and $BD = DC$, by SAS Similarity, $\triangle EDC \sim \triangle ACB$. From the similarity, $AB = 2 \cdot ED$ and $AF = \tfrac23 \cdot AB = \tfrac43 \cdot ED$.


Thus, $\frac{ED}{AF} = \frac{EP}{PF} = \frac34$, so $m+n = \boxed{7}$.

Solution 2 (credit to jeffisepic)

Since $D$ is the midpoint of $CB$, $CD = DB$. By SAS Congruency, $\triangle ACD \cong \triangle ABD$, so $\angle CAD = \angle BAD$.


By the Angle Bisector Theorem, $\tfrac{AE}{EP} = \tfrac{AF}{FP}$. We know that $AC = AB, AE = \tfrac12 \cdot AC, AF = \tfrac23 \cdot AB$, so $\tfrac{EP}{FP} = \tfrac{3}{4}$. Thus, $m+n = \boxed{7}$.

See Also

2006 iTest (Problems)
Preceded by:
Problem 29
Followed by:
Problem 31
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