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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 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>\displaystyle BX = 105 - EX = 105 - \left[\frac{287 - PQ}{2}\right]</math>. Also, <math>\displaystyle BX = BY</math>. <math>FY = 364 - BY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right]</math>.  
+
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>\displaystyle 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 theorem]], we have that <math>\displaystyle FY = PQ + QF</math>. Solve for <math>\displaystyle PQ = FY - QF</math>. Also, <math>\displaystyle QF = EP = EX</math>, so <math>\displaystyle PQ = FY - EX</math>. Since <math>\displaystyle BX = BY</math>, this can become <math>\displastyle PQ = FY - EX + (BY - BX)</math><math> = \left(FY + BY\right) - \left(EX + EY\right) = FB - EB</math>. Substituting in their values, the answer is <math>364 - 105 = 259</math>.
+
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===
 +
 
 +
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:
 +
 
 +
<cmath>[BEF] = [DEF] = 11 466</cmath>.
 +
 
 +
So the inradius of the triangle-type things is <math>\frac {637}{21}</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 ==
Line 19: Line 50:
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 10:05, 7 June 2022

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 $c$ being 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 $PQ$ we desired to find is just $c - a$. What is $b$ then? $b = 13$. And $c$ is $39$. Therefore the answer is $37$... $NOT.$

Multiply by $7$ back again (I hope you remembered to write this in $huge$ letters on top of the scrap paper!), we actually get $259$.

Solution 5

Scaling everything by 7, we have that $AE = 12, AB = 9, BF = 52$. Note that if the perpendicular of $F$ dropped down to $ED$ is $X$, then $EX = 52-12 = 40$. But $FX = 9$ and so we have a $9-40-41$ right triangle with $EFX$ meaning $EF = 41$. Now, by symmetry, we know that $EP = QF = a$ meaning $PF = 41-a$. If the tangent of the circle inscribed in $BEF$ is tangent to $BE$ at $Y$, then if $BY = b$ we have a system of equations. $a+b = 15, b+41-a = 52$. We can then solve for $a$, and since $PQ = 41-2a$, the rest follows.

See also

2007 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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