Difference between revisions of "1985 AHSME Problems/Problem 3"
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Revision as of 11:59, 5 July 2013
Problem
In right 
with legs 
and 
, arcs of circles are drawn, one with center 
and radius , the other with center 
and radius . They intersect the hypotenuse at 
and 
. Then, 
has length:
![[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(12,7), C=(12,0), M=12*dir(A--B), N=B+B.y*dir(B--A); real r=degrees(B); draw(A--B--C--cycle^^Arc(A,12,0,r)^^Arc(B,B.y,180+r,270)); pair point=incenter(A,B,C); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$M$", M, dir(point--M)); label("$N$", N, dir(point--N)); label("$12$", (6,0), S); label("$5$", (12,3.5), E);[/asy]](http://latex.artofproblemsolving.com/d/4/e/d4e3f03cc4c6e6f3527229fb934a9a015730335e.png)
Solution
First of all, from the Pythagorean Theorem, 
. Also, since 
and 
are radii of the same circle, 
. Therefore, 
. Also, since 
and 
are radii of the same circle, 
. We therefore have 
.
See Also
| 1985 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
