Difference between revisions of "2007 AIME II Problems/Problem 9"
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Several [[Pythagorean triple]]s exist amongst the numbers given. <math>BE = DF = \sqrt{63^2 + 84^2} = 21\sqrt{3^2 + 4^2} = 105</math>. Also, the length of <math>EF = \sqrt{63^2 + (448 - 2\cdot84)^2} = 7\sqrt{9^2 + 40^2} = 287</math>. | Several [[Pythagorean triple]]s exist amongst the numbers given. <math>BE = DF = \sqrt{63^2 + 84^2} = 21\sqrt{3^2 + 4^2} = 105</math>. Also, the length of <math>EF = \sqrt{63^2 + (448 - 2\cdot84)^2} = 7\sqrt{9^2 + 40^2} = 287</math>. | ||
− | Use the [[Two Tangent | + | Use the [[Two Tangent Theorem]] on <math>\triangle BEF</math>. Since both circles are inscribed in congruent triangles, they are congruent; therefore, <math>EP = FQ = \frac{287 - PQ}{2}</math>. By the Two Tangent theorem, note that <math>EP = EX = \frac{287 - PQ}{2}</math>, making <math>BX = 105 - EX = 105 - \left[\frac{287 - PQ}{2}\right]</math>. Also, <math>BX = BY</math>. <math>FY = 364 - BY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right]</math>. |
− | Finally, <math>FP = FY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right] = \frac{805 - PQ}{2}</math>. Also, <math>FP = FQ + PQ = \frac{287 - PQ}{2} + PQ</math>. Equating, we see that <math>\frac{805 - PQ}{2} = \frac{287 + PQ}{2}</math>, so <math>PQ = 259</math>. | + | Finally, <math>FP = FY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right] = \frac{805 - PQ}{2}</math>. Also, <math>FP = FQ + PQ = \frac{287 - PQ}{2} + PQ</math>. Equating, we see that <math>\frac{805 - PQ}{2} = \frac{287 + PQ}{2}</math>, so <math>PQ = \boxed{259}</math>. |
=== Solution 2 === | === Solution 2 === | ||
− | By the [[Two Tangent | + | By the [[Two Tangent Theorem]], we have that <math>FY = PQ + QF</math>. Solve for <math>PQ = FY - QF</math>. Also, <math>QF = EP = EX</math>, so <math>PQ = FY - EX</math>. Since <math>BX = BY</math>, this can become <math>PQ = FY - EX + (BY - BX)</math><math> = \left(FY + BY\right) - \left(EX + BX\right) = FB - EB</math>. Substituting in their values, the answer is <math>364 - 105 = 259</math>. |
===Solution 3=== | ===Solution 3=== | ||
− | Call the incenter of <math>\triangle BEF</math> <math>O_1</math> and the incenter of <math>\triangle | + | Call the incenter of <math>\triangle BEF</math> <math>O_1</math> and the incenter of <math>\triangle DFE</math> <math>O_2</math>. Draw triangles <math>\triangle O_1PQ,\triangle PQO_2</math>. |
Drawing <math>BE</math>, We find that <math>BE = \sqrt {63^2 + 84^2} = 105</math>. Applying the same thing for <math>F</math>, we find that <math>FD = 105</math> as well. Draw a line through <math>E,F</math> parallel to the sides of the rectangle, to intersect the opposite side at <math>E_1,F_1</math> respectively. Drawing <math>\triangle EE_1F</math> and <math>FF_1E</math>, we can find that <math>EF = \sqrt {63^2 + 280^2} = 287</math>. We then use Heron's formula to get: | Drawing <math>BE</math>, We find that <math>BE = \sqrt {63^2 + 84^2} = 105</math>. Applying the same thing for <math>F</math>, we find that <math>FD = 105</math> as well. Draw a line through <math>E,F</math> parallel to the sides of the rectangle, to intersect the opposite side at <math>E_1,F_1</math> respectively. Drawing <math>\triangle EE_1F</math> and <math>FF_1E</math>, we can find that <math>EF = \sqrt {63^2 + 280^2} = 287</math>. We then use Heron's formula to get: | ||
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Now, we just have to find <math>O_1Q = O_2P</math>, which can be done with simple subtraction, and then we can use the [[Pythagorean Theorem]] to find <math>PQ</math>. | Now, we just have to find <math>O_1Q = O_2P</math>, which can be done with simple subtraction, and then we can use the [[Pythagorean Theorem]] to find <math>PQ</math>. | ||
− | + | ==Solution 4== | |
+ | |||
+ | Why not first divide everything by its greatest common factor, <math>7</math>? Then we're left with much simpler numbers which saves a lot of time. In the end, we will multiply by <math>7</math>. | ||
+ | |||
+ | From there, we draw the same diagram as above (with smaller numbers). We soon find that the longest side of both triangles is 52 (64 - 12). That means: | ||
+ | |||
+ | <math>A = rs</math> indicating <math>26(9)=r(54)</math> so <math>r = 13/3</math>. | ||
+ | |||
+ | Now, we can start applying the equivalent tangents. Calling them <math>a</math>, <math>b</math>, and <math>c</math> (with <math>c</math> being the longest and a being the shortest), | ||
+ | |||
+ | <math>a+b+c</math> is the semi perimeter or <math>54</math>. And since the longest side (which has <math>b+c</math>) is <math>52</math>, <math>a=2</math>. | ||
+ | |||
+ | Note that the distance <math>PQ</math> we desired to find is just <math>c - a</math>. What is <math>b</math> then? <math>b = 13</math>. And <math>c</math> is <math>39</math>. Therefore the answer is <math>37</math>... <math>NOT.</math> | ||
+ | |||
+ | Multiply by <math>7</math> back again (I hope you remembered to write this in <math>huge</math> letters on top of the scrap paper!), we actually get <math>259</math>. | ||
+ | |||
+ | ==Solution 5== | ||
+ | Scaling everything by 7, we have that <math>AE = 12, AB = 9, BF = 52</math>. Note that if the perpendicular of <math>F</math> dropped down to <math>ED</math> is <math>X</math>, then <math>EX = 52-12 = 40</math>. But <math>FX = 9</math> and so we have a <math>9-40-41</math> right triangle with <math>EFX</math> meaning <math>EF = 41</math>. Now, by symmetry, we know that <math>EP = QF = a</math> meaning <math>PF = 41-a</math>. If the tangent of the circle inscribed in <math>BEF</math> is tangent to <math>BE</math> at <math>Y</math>, then if <math>BY = b</math> we have a system of equations. <math>a+b = 15, b+41-a = 52</math>. We can then solve for <math>a</math>, and since <math>PQ = 41-2a</math>, the rest follows. | ||
== See also == | == See also == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 11:05, 7 June 2022
Contents
Problem
Rectangle is given with
and
Points
and
lie on
and
respectively, such that
The inscribed circle of triangle
is tangent to
at point
and the inscribed circle of triangle
is tangent to
at point
Find
Solution
Solution 1
Several Pythagorean triples exist amongst the numbers given. . Also, the length of
.
Use the Two Tangent Theorem on . Since both circles are inscribed in congruent triangles, they are congruent; therefore,
. By the Two Tangent theorem, note that
, making
. Also,
.
.
Finally, . Also,
. Equating, we see that
, so
.
Solution 2
By the Two Tangent Theorem, we have that . Solve for
. Also,
, so
. Since
, this can become
. Substituting in their values, the answer is
.
Solution 3
Call the incenter of
and the incenter of
. Draw triangles
.
Drawing , We find that
. Applying the same thing for
, we find that
as well. Draw a line through
parallel to the sides of the rectangle, to intersect the opposite side at
respectively. Drawing
and
, we can find that
. We then use Heron's formula to get:
.
So the inradius of the triangle-type things is .
Now, we just have to find , which can be done with simple subtraction, and then we can use the Pythagorean Theorem to find
.
Solution 4
Why not first divide everything by its greatest common factor, ? Then we're left with much simpler numbers which saves a lot of time. In the end, we will multiply by
.
From there, we draw the same diagram as above (with smaller numbers). We soon find that the longest side of both triangles is 52 (64 - 12). That means:
indicating
so
.
Now, we can start applying the equivalent tangents. Calling them ,
, and
(with
being the longest and a being the shortest),
is the semi perimeter or
. And since the longest side (which has
) is
,
.
Note that the distance we desired to find is just
. What is
then?
. And
is
. Therefore the answer is
...
Multiply by back again (I hope you remembered to write this in
letters on top of the scrap paper!), we actually get
.
Solution 5
Scaling everything by 7, we have that . Note that if the perpendicular of
dropped down to
is
, then
. But
and so we have a
right triangle with
meaning
. Now, by symmetry, we know that
meaning
. If the tangent of the circle inscribed in
is tangent to
at
, then if
we have a system of equations.
. We can then solve for
, and since
, the rest follows.
See also
2007 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.