Difference between revisions of "2008 AIME II Problems/Problem 5"
m (→Solution 2) |
m (→Solution 2) |
||
Line 59: | Line 59: | ||
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); | ||
Line 65: | Line 65: | ||
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 .
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
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); (Error compiling LaTeX. 228aa9a09ec6a18b4a0ab11366b7b8f2ec44b852.asy: 27.6: no matching function 'label(string, pair[], pair)')
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 |