Difference between revisions of "Mock AIME 1 2007-2008 Problems/Problem 11"
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== Solution == | == Solution == | ||
− | <center>< | + | <center><asy> |
− | size(150); defaultpen(linewidth(0.8)); import markers | + | size(150); defaultpen(linewidth(0.8)); import markers; |
pair B = (0,0), C = (25,0), A = (578/50,19.8838); | pair B = (0,0), C = (25,0), A = (578/50,19.8838); | ||
draw(A--B--C--cycle); | draw(A--B--C--cycle); | ||
− | label(" | + | label("$B$",B,SW); label("$C$",C,SE); label("$A$",A,N); |
pair D = (13,0), E = (11*A + 13*C)/24, F = (12*B + 11*A)/23; | pair D = (13,0), E = (11*A + 13*C)/24, F = (12*B + 11*A)/23; | ||
draw(D--E--F--cycle); | draw(D--E--F--cycle); | ||
− | label(" | + | label("$D$",D,dir(-90)); |
− | label(" | + | label("$E$",E,dir(0)); |
− | label(" | + | label("$F$",F,dir(180)); |
draw(A--E,StickIntervalMarker(1,3,size=6));draw(B--D,StickIntervalMarker(1,3,size=6)); | draw(A--E,StickIntervalMarker(1,3,size=6));draw(B--D,StickIntervalMarker(1,3,size=6)); |
Revision as of 17:09, 18 May 2012
Problem
is inscribed inside such that lie on , respectively. The circumcircles of have centers , respectively. Also, , and . The length of can be written in the form , where and are relatively prime integers. Find .
Solution
From adjacent sides, the following relationships can be derived:
$
Since , and , . Thus, . . Thus, . Thus, the answer is .
See also
Mock AIME 1 2007-2008 (Problems, Source) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |