# 2008 AIME II Problems/Problem 5

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## Problem 5

In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$, let $BC = 1000$ and $AD = 2008$. Let $\angle A = 37^\circ$, $\angle D = 53^\circ$, and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$, respectively. Find the length $MN$.

## Solution

### Solution 1

Extend $\overline{AD}$ and $\overline{BC}$ to meet at a point $E$. Then $\angle AED = 180 - 53 - 37 = 90^{\circ}$.

$[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); draw(A--B--C--D--cycle); draw(B--E--C,dashed); draw(M[0]--N); draw(N--E,dashed); draw(rightanglemark(D,E,A,2)); picture p = new picture; draw(p,Circle(N,r),dashed+linewidth(0.5)); clip(p,A--D--D+(0,20)--A+(0,20)--cycle); add(p); label("$$A$$",A,SW); label("$$B$$",B,NW); label("$$C$$",C,NE); label("$$D$$",D,SE); label("$$E$$",E,NE); label("$$M$$",M[0],SW); label("$$N$$",N,S); label("$$1004$$",(N+D)/2,S); label("$$500$$",(M[0]+C)/2,S); [/asy]$

Since $\overline{BC} \parallel \overline{AD}$, then $BC$ and $AD$ are homothetic with respect to point $E$ by a ratio of $\frac{BC}{AD} = \frac{125}{251}$. Since the homothety carries the midpoint of $\overline{BC}$, $M$, to the midpoint of $\overline{AD}$, which is $N$, then $E,M,N$ are collinear.

As $\angle AED = 90^{\circ}$, note that the midpoint of $\overline{AD}$, $N$, is the center of the circumcircle of $\triangle AED$. It follows that $$NE = ND = \frac{AD}{2} = 1004.$$ Since $\triangle EMC \sim \triangle END$, we have that $$\frac{EM}{EM + MN} = \frac{MC}{ND} = \frac{125}{251} \Longrightarrow MN = \frac{125}{126}EM.$$ Substituting, $1004 = NE = EM + MN = \frac{251}{126}MN$, and $MN = \boxed{504}$.

### Solution 2

$[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label("$$A$$",A,SW); label("$$B$$",B,NW); label("$$C$$",C,NE); label("$$D$$",D,NE); label("$$F$$",F,S); label("$$G$$",G,SW); label("$$M$$",M[0],SW); label("$$N$$",N,S); label("$$H$$",H,S); label("$$x$$",(A+F)/2,S); label("$$h$$",(B+F)/2,W); label("$$h$$",(C+G)/2,W); label("$$1000$$",(B+C)/2,NE); label("$$1008-x$$",(G+D)/2,S); [/asy]$

Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\overline{AD}$, respectively. Let $x = AF$, so $FG = 1000$ and $DG = 2008 - x - 1000 = 1008 - x$. Also, let $h = BF = CG = HM$.

By AA~, we have that $\triangle AFB \sim \triangle CGD$, and so $$\frac{BF}{AF} = \frac {CG}{DG} \Longleftrightarrow \frac{h}{x} = \frac{1008-x}{h} \Longrightarrow h^2 + x^2 - 1008x = 0$$.

Since $AH = 500+x$ and $AN = 1004$, it follows that $HN = AN - AH = 504 - x$. By the Pythagorean Theorem on $\triangle MON$, $$MN^{2} = (504 - x)^2 + h^2 = 504^2 + h^2 + x^2 - 1008x = 504^2$$ and the answer is $MN = 504$.