Difference between revisions of "2008 AIME II Problems/Problem 5"
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label("\(N\)",N,S); | label("\(N\)",N,S); | ||
label("\(H\)",H,S); | label("\(H\)",H,S); | ||
− | label("\(x\)",(N+H)/2, | + | label("\(x\)",(N+H)/2,N); |
label("\(h\)",(B+F)/2,W); | label("\(h\)",(B+F)/2,W); | ||
label("\(h\)",(C+G)/2,W); | label("\(h\)",(C+G)/2,W); | ||
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label("\(504-x\)",(G+D)/2,S); | label("\(504-x\)",(G+D)/2,S); | ||
label("\(504+x\)",(A+F)/2,S); | label("\(504+x\)",(A+F)/2,S); | ||
+ | label("\(h\)",(M+N)/2,W); | ||
</asy></center> | </asy></center> | ||
Let <math>F,G,H</math> be the feet of the [[perpendicular]]s from <math>B,C,M</math> onto <math>\overline{AD}</math>, respectively. Let <math>x = NH</math>, so <math>DG = 1004 - 500 - x = 504 - x</math> and <math>AF = 1004 - (504 - x) = 504 + x</math>. Also, let <math>h = BF = CG = HM</math>. | Let <math>F,G,H</math> be the feet of the [[perpendicular]]s from <math>B,C,M</math> onto <math>\overline{AD}</math>, respectively. Let <math>x = NH</math>, so <math>DG = 1004 - 500 - x = 504 - x</math> and <math>AF = 1004 - (504 - x) = 504 + x</math>. Also, let <math>h = BF = CG = HM</math>. |
Revision as of 23:33, 3 April 2008
Problem 5
In trapezoid with
, let
and
. Let
,
, and
and
be the midpoints of
and
, respectively. Find the length
.
Solution
Solution 1
Extend and
to meet at a point
. Then
.
![[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]](http://latex.artofproblemsolving.com/a/d/e/ade36a6d8b6d97f88722facc76700d34501ddae8.png)
Since , then
and
are homothetic with respect to point
by a ratio of
. Since the homothety carries the midpoint of
,
, to the midpoint of
, which is
, then
are collinear.
As , note that the midpoint of
,
, is the center of the circumcircle of
. We can do the same with the circumcircle about
and
(or we could apply the homothety to find
in terms of
). It follows that
Thus
.
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,N); 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+N)/2,W); [/asy]](http://latex.artofproblemsolving.com/f/3/d/f3db0d62772edda4f212f4c04c8e16356f4dda62.png)
Let be the feet of the perpendiculars from
onto
, respectively. Let
, so
and
. Also, let
.
By AA~, we have that , and so
By the Pythagorean Theorem on ,
so
.
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 |