Difference between revisions of "1993 AIME Problems/Problem 15"
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{{AIME box|year=1993|num-b=14|after=Last question}} | {{AIME box|year=1993|num-b=14|after=Last question}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
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Revision as of 19:26, 4 July 2013
Problem
Let 
be an altitude of 
. Let 
and 
be the points where the circles inscribed in the triangles 
and 
are tangent to . If 
, 
, and 
, then 
can be expressed as 
, where 
and 
are relatively prime integers. Find 
.
Solution
![[asy] unitsize(48); pair A,B,C,H; A=(8,0); B=origin; C=(3,4); H=(3,0); draw(A--B--C--cycle); draw(C--H); label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,N); label("$H$",H,NE); draw(circle((2,1),1)); pair [] x=intersectionpoints(C--H,circle((2,1),1)); dot(x[0]); label("$S$",x[0],SW); draw(circle((4.29843788128,1.29843788128),1.29843788128)); pair [] y=intersectionpoints(C--H,circle((4.29843788128,1.29843788128),1.29843788128)); dot(y[0]); label("$R$",y[0],NE); label("$1993$",(1.5,2),NW); label("$1994$",(5.5,2),NE); label("$1995$",(4,0),S); [/asy]](http://latex.artofproblemsolving.com/b/8/b/b8b732c641481c3b1635527f52f3125316ac3f3d.png)
From the Pythagorean Theorem, 
, and 
. Subtracting those two equations yields 
. After simplification, we see that 
, or 
. Note that 
. Therefore we have that 
. Therefore 
.
Now note that 
, 
, and 
. Therefore we have
.
Plugging in 
and simplifying, we have 
.
See also
| 1993 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Last question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
