Difference between revisions of "2008 AIME II Problems/Problem 5"

Revision as of 22:40, 3 April 2008

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 $\triangle AED$ . We can do the same with the circumcircle about $\triangle BEC$ and $M$ (or we could apply the homothety to find $ME$ in terms of $NE$ ). It follows that $NE = ND = \frac {AD}{2} = 1004, \quad ME = MC = \frac {BC}{2} = 500$ Thus $MN = NE - ME = \boxed{504}$ .

Solution 2

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("\(x\)",(N+H)/2,S);
label("\(h\)",(B+F)/2,W);
label("\(h\)",(C+G)/2,W);
label("\(1000\)",(B+C)/2,NE);
label("\(504-x\)",(G+D)/2,S);
label("\(504+x\)",(A+F)/2,S);
label("\(h\)",(M+H)/2,W);
 (Error making remote request. Unknown error_msg)

Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\overline{AD}$ , respectively. Let $x = NH$ , so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (504 - x) = 504 + x$ . Also, let $h = BF = CG = HM$ .

By AA~, we have that $\triangle AFB \sim \triangle CGD$ , and so $\frac{BF}{AF} = \frac {DG}{CG} \Longleftrightarrow \frac{h}{504+x} = \frac{504-x}{h} \Longrightarrow h^2 = (504 + x)(504 - x)\Longrightarrow x^2 + h^2 = 504^2.$

By the Pythagorean Theorem on $\triangle MHN$ , $MN^{2} = x^2 + h^2 = 504^2,$ so $MN = \boxed{504}$ .

@@ Line 70: / Line 70: @@
 By AA~, we have that <math>\triangle AFB \sim \triangle CGD</math>, and so
-<cmath>\frac{BF}{AF} = \frac {CG}{DG} \Longleftrightarrow \frac{h}{504+x} = \frac{504-x}{h} \Longrightarrow h^2 = (504 + x)(504 - x)\Longrightarrow x^2 + h^2 = 504^2.</cmath>
+<cmath>\frac{BF}{AF} = \frac {DG}{CG} \Longleftrightarrow \frac{h}{504+x} = \frac{504-x}{h} \Longrightarrow h^2 = (504 + x)(504 - x)\Longrightarrow x^2 + h^2 = 504^2.</cmath>
 By the [[Pythagorean Theorem]] on <math>\triangle MHN</math>,

2008 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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