Difference between revisions of "2001 AIME II Problems/Problem 7"
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===Note=== | ===Note=== | ||
− | The problem can be reduced to a <math>3-4-5</math> triangle for the initial calculations, where <math>C_2</math> is calculated to be (<math>\frac{5}{2}, \frac{1}{2})</math>, and <math>C_3</math> is calculated to be (<math>\frac{1}{3}, \frac{7}{3})</math>. After we find the incenters the points can be scaled up by a factor of 30 for the final distance calculation. | + | The problem can be reduced to a <math>3-4-5</math> triangle for the initial calculations, where <math>C_2</math> is calculated to be (<math>\frac{5}{2}, \frac{1}{2})</math>, and <math>C_3</math> is calculated to be (<math>\frac{1}{3}, \frac{7}{3})</math>. After we find the incenters the points can be scaled up by a factor of <math>30</math> for the final distance calculation. |
+ | |||
+ | === Solution 4 (easy but hard to see) === | ||
+ | We can calculate the inradius of each triangle as with the previous solutions. Now, notice that the hexagon <math>PSTVU</math> is a square with its corner cut off. We literally complete the square and mark the last corner as point X. Now, construct right triangle with <math>C_3C_2</math> as its hypotenuse. The right angle will be at point <math>Y</math>. We will now find the length of each leg and use the Pythagorean Theorem to compute the desired length. We see that the length of <math>C_3Y</math> is <math>50 + 15 = 65</math>, as seen by the inradius of <math>C_2</math> and <math>10</math> less than the square's side length. <math>C_2Y</math> is <math>45 + 10 = 55</math>, which is <math>15</math> less than the square plus the inradius of <math>C_3</math>. Our final answer is <math>\sqrt{65^2 + 55^2} = \sqrt{7250} = \sqrt{(10)(725)} \rightarrow \boxed{725}</math>. | ||
== See also == | == See also == |
Latest revision as of 02:15, 26 December 2022
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
for the final distance calculation.
Solution 4 (easy but hard to see)
We can calculate the inradius of each triangle as with the previous solutions. Now, notice that the hexagon is a square with its corner cut off. We literally complete the square and mark the last corner as point X. Now, construct right triangle with
as its hypotenuse. The right angle will be at point
. We will now find the length of each leg and use the Pythagorean Theorem to compute the desired length. We see that the length of
is
, as seen by the inradius of
and
less than the square's side length.
is
, which is
less than the square plus the inradius of
. Our final answer is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.