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Difference between revisions of "2007 AIME II Problems/Problem 9"

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==Solution 4==
 
==Solution 4==
  
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Why not first divide everything by its greatest common factor, 7? Then we're left with much simpler numbers which saves a lot of time. In the end, we will multiply by 7.
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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:
 
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:
  
−
A = rs indicating 26(9)=r(54) so r = 13/3.
+
<math>A = rs</math> indicating <math>26(9)=r(54)</math> so<math> r = 13/3</math>.
  
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Now, we can start applying the equivalent tangents. Calling them a, b, and c (with c being the longest and a being the shortest),
+
Now, we can start applying the equivalent tangents. Calling them <math>a</math>, <math>b</math>, and <math>c</math> (with <math>c 4being the longest and a being the shortest),
  
−
a+b+c is the semi perimeter or 54. And since the longest side (which has b+c) is 52, a=2.
+
</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$.
  
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Note that the distance PQ we desired to find is just c - a. What is b then? b = 13. And c is 39. Therefore the answer is 37.
 
 
== See also ==
 
== See also ==
 
{{AIME box|year=2007|n=II|num-b=8|num-a=10}}
 
{{AIME box|year=2007|n=II|num-b=8|num-a=10}}

Revision as of 20:20, 9 March 2017

Problem

Rectangle $ABCD$ is given with $AB=63$ and $BC=448.$ Points $E$ and $F$ lie on $AD$ and $BC$ respectively, such that $AE=CF=84.$ The inscribed circle of triangle $BEF$ is tangent to $EF$ at point $P,$ and the inscribed circle of triangle $DEF$ is tangent to $EF$ at point $Q.$ Find $PQ.$

Solution

2007 AIME II-9.png

Solution 1

Several Pythagorean triples exist amongst the numbers given. $BE = DF = \sqrt{63^2 + 84^2} = 21\sqrt{3^2 + 4^2} = 105$. Also, the length of $EF = \sqrt{63^2 + (448 - 2\cdot84)^2} = 7\sqrt{9^2 + 40^2} = 287$.

Use the Two Tangent Theorem on $\triangle BEF$. Since both circles are inscribed in congruent triangles, they are congruent; therefore, $EP = FQ = \frac{287 - PQ}{2}$. By the Two Tangent theorem, note that $EP = EX = \frac{287 - PQ}{2}$, making $BX = 105 - EX = 105 - \left[\frac{287 - PQ}{2}\right]$. Also, $BX = BY$. $FY = 364 - BY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right]$.

Finally, $FP = FY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right] = \frac{805 - PQ}{2}$. Also, $FP = FQ + PQ = \frac{287 - PQ}{2} + PQ$. Equating, we see that $\frac{805 - PQ}{2} = \frac{287 + PQ}{2}$, so $PQ = \boxed{259}$.

Solution 2

By the Two Tangent Theorem, we have that $FY = PQ + QF$. Solve for $PQ = FY - QF$. Also, $QF = EP = EX$, so $PQ = FY - EX$. Since $BX = BY$, this can become $PQ = FY - EX + (BY - BX)$$= \left(FY + BY\right) - \left(EX + BX\right) = FB - EB$. Substituting in their values, the answer is $364 - 105 = 259$.

Solution 3

Call the incenter of $\triangle BEF$ $O_1$ and the incenter of $\triangle DFE$ $O_2$. Draw triangles $\triangle O_1PQ,\triangle PQO_2$.

Drawing $BE$, We find that $BE = \sqrt {63^2 + 84^2} = 105$. Applying the same thing for $F$, we find that $FD = 105$ as well. Draw a line through $E,F$ parallel to the sides of the rectangle, to intersect the opposite side at $E_1,F_1$ respectively. Drawing $\triangle EE_1F$ and $FF_1E$, we can find that $EF = \sqrt {63^2 + 280^2} = 287$. We then use Heron's formula to get:

\[[BEF] = [DEF] = 11 466\].

So the inradius of the triangle-type things is $\frac {637}{21}$.

Now, we just have to find $O_1Q = O_2P$, which can be done with simple subtraction, and then we can use the Pythagorean Theorem to find $PQ$.

Solution 4

Why not first divide everything by its greatest common factor, $7$? Then we're left with much simpler numbers which saves a lot of time. In the end, we will multiply by $7$.

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:

$A = rs$ indicating $26(9)=r(54)$ so$r = 13/3$.

Now, we can start applying the equivalent tangents. Calling them $a$, $b$, and $c$ (with $c 4being the longest and a being the shortest),$a+b+c$is the semi perimeter or$54$. And since the longest side (which has$b+c$) is$52$,$a=2$.

Note that the distance$ (Error compiling LaTeX. Unknown error_msg)PQ$we desired to find is just$c - a$. What is$b$then?$ b = 13$. And$c$is$39$. Therefore the answer is$37$.

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

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