Difference between revisions of "2001 AIME II Problems/Problem 7"
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− | Let <math>P = (0,0)</math> be at the origin. Using the formula <math>A = rs</math> on <math>\triangle PQR</math>, where <math>r_{1}</math> is the [[inradius]] (similarly define <math>r_2, r_3</math> to be the radii of <math>C_2, C_3</math>), <math>s = \frac{PQ + QR + RP}{2} = 180</math> is the [[semiperimeter]], and <math>A = \frac 12 bh = 5400</math> is the area, we find <math>r_{1} = \frac As = 30</math>. Thus <math>ST, UV</math> lie respectively on the lines <math>y = 60, x = 60</math>, and so <math>RS = 60, UQ = 30</math>. | + | Let <math>P = (0,0)</math> be at the origin. Using the formula <math>A = rs</math> on <math>\triangle PQR</math>, where <math>r_{1}</math> is the [[inradius]] (similarly define <math>r_2, r_3</math> to be the radii of <math>C_2, C_3</math>), <math>s = \frac{PQ + QR + RP}{2} = 180</math> is the [[semiperimeter]], and <math>A = \frac 12 bh = 5400</math> is the area, we find <math>r_{1} = \frac As = 30</math>. Or, you could find the inradius directly by using the formula <math>\frac{a+b-c}{2}</math>, where <math>a</math> and <math>b</math> are the legs of the right triangle and <math>c</math> is the hypotenuse. (Can you see why? Plus, use this formula ''only for right triangles'' because these are actually the only cases.) Thus <math>ST, UV</math> lie respectively on the lines <math>y = 60, x = 60</math>, and so <math>RS = 60, UQ = 30</math>. |
Note that <math>\triangle PQR \sim \triangle STR \sim \triangle UQV</math>. Since the ratio of corresponding lengths of similar figures are the same, we have | Note that <math>\triangle PQR \sim \triangle STR \sim \triangle UQV</math>. Since the ratio of corresponding lengths of similar figures are the same, we have |
Revision as of 21:57, 11 February 2012
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
[hide]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, you could find the inradius directly by using the formula
, where
and
are the legs of the right triangle and
is the hypotenuse. (Can you see why? Plus, use this formula only for right triangles because these are actually the only cases.) 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
.
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 |