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

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Since <math>\overline{BC} \parallel \overline{AD}</math>, then <math>BC</math> and <math>AD</math> are [[homothecy|homothetic]] with respect to point <math>E</math> by a ratio of <math>\frac{BC}{AD} = \frac{125}{251}</math>. Since the homothety carries the midpoint of <math>\overline{BC}</math>, <math>M</math>, to the midpoint of <math>\overline{AD}</math>, which is <math>N</math>, then <math>E,M,N</math> are [[collinear]].  
 
Since <math>\overline{BC} \parallel \overline{AD}</math>, then <math>BC</math> and <math>AD</math> are [[homothecy|homothetic]] with respect to point <math>E</math> by a ratio of <math>\frac{BC}{AD} = \frac{125}{251}</math>. Since the homothety carries the midpoint of <math>\overline{BC}</math>, <math>M</math>, to the midpoint of <math>\overline{AD}</math>, which is <math>N</math>, then <math>E,M,N</math> are [[collinear]].  
  
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As <math>\angle AED = 90^{\circ}</math>, note that the midpoint of <math>\overline{AD}</math>, <math>N</math>, is the center of the [[circumcircle]] of <math>\triangle AED</math>. It follows that <cmath>NE = ND = \frac{AD}{2} = 1004.</cmath> Since <math>\triangle EMC \sim \triangle END</math>, we have that <cmath>\frac{EM}{EM + MN} = \frac{MC}{ND} = \frac{125}{251} \Longrightarrow MN = \frac{125}{126}EM.</cmath> Substituting, <math>1004 = NE = EM + MN = \frac{251}{126}MN</math>, and <math>MN = \boxed{504}</math>.
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As <math>\angle AED = 90^{\circ}</math>, note that the midpoint of <math>\overline{AD}</math>, <math>N</math>, is the center of the [[circumcircle]] of <math>\triangle AED</math>. We can do the same with the circumcircle about <math>\triangle BEC</math> and <math>M</math> (or we could apply the homothety to find <math>ME</math> in terms of <math>NE</math>). It follows that
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<cmath>
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NE = ND = \frac {AD}{2} = 1004, \quad ME = MC = \frac {BC}{2} = 500
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</cmath>
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Thus <math>MN = NE - ME = \boxed{504}</math>.
  
 
=== Solution 2 ===
 
=== Solution 2 ===

Revision as of 18:17, 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

[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 $MN = 504$.

See also

2008 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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