2008 AIME II Problems/Problem 5

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]$

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 $MN = NE - ME = \boxed{504}$ .

For purposes of rigor we will show that $E,M,N$ are collinear. 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.

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$",(N+H)/2+(0,1),S); label("$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); [/asy]$

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 $AF = 1004 - (500 - x) = 504 + x$ . Also, let $h = BF = CG = HM$ .

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

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

Solution 3

If you drop perpendiculars from $B$ and $C$ to $AD$ , and call the points if you drop perpendiculars from $B$ and $C$ to $\overline{AD}$ and call the points where they meet $\overline{AD}$ , $E$ and $F$ respectively and call $FD = x$ and $EA = 1008-x$ , then you can solve an equation in tangents. Since $\angle{A} = 37$ and $\angle{D} = 53$ , you can solve the equation [by cross-multiplication]:

$\begin{align*}\tan{37}\times (1008-x) &= \tan{53} \times x\\ \frac{(1008-x)}{x} &= \frac{\tan{53}}{\tan{37}} = \frac{\sin{53}}{\cos{53}} \times\frac{\sin{37}}{\cos{37}}\end{align*}$

However, we know that $\cos{90-x} = \sin{x}$ and $\sin{90-x} = \cos{x}$ are co-functions. Applying this,

$\begin{align*}\frac{(1008-x)}{x} &= \frac{\sin^2{53}}{\cos^2{53}} \\ x\sin^2{53} &= 1008\cos^2{53} - x\cos^2{53}\\ x(\sin^2{53} + \cos^2{53}) &= 1008\cos^2{53}\\ x = 1008\cos^2{53} &\Longrightarrow 1008-x = 1008\sin^2{53} \end{align*}$ Now, if we can find $1004 - (EA + 500)$ , and the height of the trapezoid, we can create a right triangle and use the Pythagorean Theorem to find $MN$ .

The leg of the right triangle along the horizontal is:

$1004 - 1008\sin^2{53} - 500 = 504 - 1008\sin^2{53}.$

Now to find the other leg of the right triangle (also the height of the trapezoid), we can simplify the following expression:

$\begin{align*}\tan{37} \times 1008 \sin^2{53} = \tan{37} \times 1008 \cos^2{37} = 1008\cos{37}\sin{37} = 504\sin74\end{align*}$

Now we used Pythagorean Theorem and get that $MN$ is equal to:

$\begin{align*}&\sqrt{(1008\sin^2{53} + 500 -1004)^2 + (504\sin{74})^2} = 504\sqrt{1-2\sin^2{53} + \sin^2{74}} \end{align*}$

However, $1-2\sin^2{53} = \cos^2{106}$ and $\sin^2{74} = \sin^2{106}$ so now we end up with:

$504\sqrt{\cos^2{106} + \sin^2{106}} =\fbox{504}.$

Solution 4

Plot the trapezoid such that $B=\left(1000\cos 37^\circ, 0\right)$ , $C=\left(0, 1000\sin 37^\circ\right)$ , $A=\left(2008\cos 37^\circ, 0\right)$ , and $D=\left(0, 2008\sin 37^\circ\right)$ .

The midpoints of the requested sides are $\left(500\cos 37^\circ, 500\sin 37^\circ\right)$ and $\left(1004\cos 37^\circ, 1004\sin 37^\circ\right)$ .

To find the distance from $M$ to $N$ , we simply apply the distance formula and the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ to get $MN=\boxed{504}$ .

Solution 5

Similar to solution 1; Notice that it forms a right triangle. Remembering that the median to the hypotenuse is simply half the length of the hypotenuse, we quickly see that the length is 2008/2-1000/2=504.

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

Page

Toolbox

Search