2001 AIME II Problems/Problem 7
Problem
Let be a right triangle with
,
, and
. Let
be the inscribed circle. Construct
with
on
and
on
, such that
is perpendicular to
and tangent to
. Construct
with
on
and
on
such that
is perpendicular to
and tangent to
. Let
be the inscribed circle of
and
the inscribed circle of
. The distance between the centers of
and
can be written as
. What is
?
Contents
Solution
Solution 1 (analytic)
![[asy] pointpen = black; pathpen = black + linewidth(0.7); pair P = (0,0), Q = (90, 0), R = (0, 120), S=(0, 60), T=(45, 60), U = (60,0), V=(60, 40), O1 = (30,30), O2 = (15, 75), O3 = (70, 10); D(MP("P",P)--MP("Q",Q)--MP("R",R,W)--cycle); D(MP("S",S,W) -- MP("T",T,NE)); D(MP("U",U) -- MP("V",V,NE)); D(O2 -- O3, rgb(0.2,0.5,0.2)+ linewidth(0.7) + linetype("4 4")); D(CR(D(O1), 30)); D(CR(D(O2), 15)); D(CR(D(O3), 10)); [/asy]](http://latex.artofproblemsolving.com/c/e/3/ce3f5916c09a4de9f57dae8e8db8b9a7638ae242.png)
Let be at the origin. Using the formula
on
, where
is the inradius (similarly define
to be the radii of
),
is the semiperimeter, and
is the area, we find
. Or, the inradius could be directly by using the formula
, where
and
are the legs of the right triangle and
is the hypotenuse. (This formula should be used only for right triangles.) Thus
lie respectively on the lines
, and so
.
Note that . Since the ratio of corresponding lengths of similar figures are the same, we have
Let the centers of be
, respectively; then by the distance formula we have
. Therefore, the answer is
.
Solution 2 (synthetic)
![[asy] pointpen = black; pathpen = black + linewidth(0.7); pair P = (0,0), Q = (90, 0), R = (0, 120), S=(0, 60), T=(45, 60), U = (60,0), V=(60, 40), O1 = (30,30), O2 = (15, 75), O3 = (70, 10); D(MP("P",P)--MP("Q",Q)--MP("R",R,W)--cycle); D(MP("S",S,W) -- MP("T",T,NE)); D(MP("U",U) -- MP("V",V,NE)); D(O2 -- O3, rgb(0.2,0.5,0.2)+ linewidth(0.7) + linetype("4 4")); D(CR(D(O1), 30)); D(CR(D(O2), 15)); D(CR(D(O3), 10)); pair A2 = IP(incircle(R,S,T), Q--R), A3 = IP(incircle(Q,U,V), Q--R); D(D(MP("A_2",A2,NE)) -- O2, linetype("4 4")+linewidth(0.6)); D(D(MP("A_3",A3,NE)) -- O3 -- foot(O3, A2, O2), linetype("4 4")+linewidth(0.6)); [/asy]](http://latex.artofproblemsolving.com/5/e/2/5e2fbf2e8f2ca6c823bafce3f6042ee372dfa534.png)
We compute as above. Let
respectively the points of tangency of
with
.
By the Two Tangent Theorem, we find that ,
. Using the similar triangles,
,
, so
. Thus
.
Solution 3
The radius of an incircle is . The area of the triangle is equal to
and the semiperimeter is equal to
. The radius, therefore, is equal to
. Thus using similar triangles the dimensions of the triangle circumscribing the circle with center
are equal to
,
, and
. The radius of the circle inscribed in this triangle with dimensions
is found using the formula mentioned at the very beginning. The radius of the incircle is equal to
.
Defining as
,
is equal to
or
. Also using similar triangles, the dimensions of the triangle circumscribing the circle with center
are equal to
,
,
or
. The radius of
by using the formula mentioned at the beginning is
. Using
as
,
is equal to
or
. Using the distance formula, the distance between
and
:
this equals
or
, thus
is
.
==Note==: The problem can be reduced to a triangle for the initial calculations, where
is calculated to be (
, and
is calculated to be (
. After we find the incenters the points can be scaled up by a factor of 30 for the final distance calculation.
See also
2001 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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