# 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 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10
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