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Difference between revisions of "2008 AMC 12A Problems/Problem 21"

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== Problem ==
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==Problem==
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[[Triangle]] <math>ABC</math> has <math>AC = 3</math>, <math>BC = 4</math>, and <math>AB = 5</math>. Point <math>D</math> is on <math>\overline{AB}</math>, and <math>\overline{CD}</math> [[bisector|bisects]] the [[right angle]]. The inscribed circles of <math>\triangle ADC</math> and <math>\triangle BCD</math> have [[radii]] <math>r_a</math> and <math>r_b</math>, respectively. What is <math>r_a/r_b</math>?
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A permutation <math>(a_1,a_2,a_3,a_4,a_5)</math> of <math>(1,2,3,4,5)</math> is <u>heavy-tailed</u> if <math>a_1 + a_2 < a_4 + a_5</math>. What is the number of heavy-tailed permutations?
  
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<math>\textbf{(A)}\ \frac {1}{28}\left(10 - \sqrt {2}\right) \qquad \textbf{(B)}\ \frac {3}{56}\left(10 - \sqrt {2}\right) \qquad \textbf{(C)}\ \frac {1}{14}\left(10 - \sqrt {2}\right) \qquad \textbf{(D)}\ \frac {5}{56}\left(10 - \sqrt {2}\right) \\
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<math>\mathrm{(A)}\ 36\qquad\mathrm{(B)}\ 40\qquad\textbf{(C)}\ 44\qquad\mathrm{(D)}\ 48\qquad\mathrm{(E)}\ 52</math>
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\textbf{(E)}\ \frac {3}{28}\left(10 - \sqrt {2}\right)</math>
 
  
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== Solution ==
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==Solution 1 (Symmetry) ==
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<center><asy>
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There are <math>5!=120</math> total permutations.
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import olympiad;
 
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size(300);
 
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defaultpen(0.8);
 
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pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429);
 
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pair O=incenter(A,C,D), P=incenter(B,C,D);
 
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picture p = new picture;
 
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draw(p,Circle(C,0.2)); draw(p,Circle(B,0.2));
 
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clip(p,B--C--D--cycle);
 
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add(p);
 
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draw(A--B--C--D--C--cycle);
 
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draw(incircle(A,C,D));
 
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draw(incircle(B,C,D));
 
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dot(O);dot(P);
 
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label("\(A\)",A,W);
 
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label("\(B\)",B,E);
 
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label("\(C\)",C,W);
 
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label("\(D\)",D,NE);
 
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label("\(O_A\)",O,W);
 
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label("\(O_B\)",P,W);
 
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label("\(3\)",(A+C)/2,W);
 
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label("\(4\)",(B+C)/2,S);
 
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label("\(\frac{15}{7}\)",(A+D)/2,NE);
 
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label("\(\frac{20}{7}\)",(B+D)/2,NE);
 
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label("\(45^{\circ}\)",(.2,.1),E);
 
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label("\(\sin \theta = \frac{3}{5}\)",B-(.2,-.1),W);
 
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</asy></center>
 
  
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By the [[Angle Bisector Theorem]],
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For every permutation <math>(a_1,a_2,a_3,a_4,a_5)</math> such that <math>a_1 + a_2 < a_4 + a_5</math>, there is exactly one permutation such that <math>a_1 + a_2 > a_4 + a_5</math>. Thus it suffices to count the permutations such that <math>a_1 + a_2 = a_4 + a_5</math>.
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<cmath>\frac{BD}{4} = \frac{5-BD}{3} \Longrightarrow BD = \frac{20}7</cmath>
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By [[Law of Sines]] on <math>\triangle BCD</math>,  
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<math>1+4=2+3</math>, <math>1+5=2+4</math>, and <math>2+5=3+4</math> are the only combinations of numbers that can satisfy <math>a_1 + a_2 = a_4 + a_5</math>.
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<cmath>\frac{BD}{\sin 45^{\circ}} = \frac{CD}{\sin \angle B} \Longrightarrow \frac{20/7}{\sqrt{2}/2} = \frac{CD}{3/5} \Longrightarrow CD=\frac{12\sqrt{2}}{7}</cmath>
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Since the area of a triangle satisfies <math>[\triangle]=rs</math>, where <math>r = </math> the [[inradius]] and <math>s =</math> the [[semiperimeter]], we have
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There are <math>3</math> combinations of numbers, <math>2</math> possibilities of which side of the equation is <math>a_1+a_2</math> and which side is <math>a_4+a_5</math>, and <math>2^2=4</math> possibilities for rearranging <math>a_1,a_2</math> and <math>a_4,a_5</math>. Thus, there are <math>3\cdot2\cdot4=24</math> permutations such that <math>a_1 + a_2 = a_4 + a_5</math>.
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<cmath>\frac{r_A}{r_B} = \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A}</cmath>
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<!--Using any of various formulas for triangle area, we find the area <math>[BCD]</math> to be
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Thus, the number of <u>heavy-tailed</u> permutations is <math>\frac{120-24}{2}=48 \Rightarrow D</math>.
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<cmath>[BCD] = \frac{1}{2} (\sin \angle CBD) \cdot (BD) \cdot (CD) = \frac 12 \cdot \frac 35 \cdot \frac{20}{7} \cdot 4 = \frac{24}{7}</cmath>
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and
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==Solution 2 (Casework)==
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<cmath>[ACD] = [ABC] - [BCD] = \frac 12 (3)(4) - \frac{24}{7} = \frac{18}{7}</cmath>-->
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We use case work on the value of <math>a_3</math>.
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Since <math>\triangle ACD</math> and <math>\triangle BCD</math> share the [[altitude]] (to <math>\overline{AB}</math>), their areas are the ratio of their bases, or <cmath>\frac{[ACD]}{[BCD]} = \frac{AD}{BD} = \frac{3}{4}</cmath>
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The semiperimeters are <math>s_A = \left({3 + \frac{15}{7} + \frac{12\sqrt{2}}{7}\right)\left/\right.2 = \frac{18+6\sqrt{2}}{7}</math> and <math>s_B = \frac{24+ 6\sqrt{2}}{7}</math>. Thus,
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Case 1: <math>a_3 = 1</math>. Since <math>a_1 + a_2 < a_4 + a_5</math>, <math>(a_1, a_2)</math> can only be a permutation of <math>(2, 3)</math> or <math>(2, 4)</math>. The values of <math>a_1</math> and <math>a_2</math>, as well as the values of <math>a_4</math> and <math>a_5</math>, are interchangeable, so this case produces a total of <math>2(2 \cdot 2) = 8</math> solutions.
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<cmath>\begin{align*}
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\frac{r_A}{r_B} &= \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A} = \frac{3}{4} \cdot \frac{(24+ 6\sqrt{2})/7}{(18+6\sqrt{2})/7} \\
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Case 2: <math>a_3 = 2</math>. Similarly, we have <math>(a_1, a_2)</math> is a permutation of <math>(1, 3)</math>, <math>(1, 4)</math>, or <math>(1, 5)</math>, which gives a total of <math>3(2 \cdot 2) = 12</math> solutions.
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&= \frac{3(4+\sqrt{2})}{4(3+\sqrt{2})} \cdot \left(\frac{3-\sqrt{2}}{3-\sqrt{2}}\right) = \frac{3}{28}(10-\sqrt{2}) \Rightarrow \mathrm{(E)}\qquad \blacksquare \end{align*}</cmath>
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Case 3: <math>a_3 = 3</math>. <math>(a_1, a_2)</math> is a permutation of <math>(1, 2)</math> or <math>(1, 4)</math>, which gives a total of <math>2(2 \cdot 2) = 8</math> solutions.
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Case 4: <math>a_3 = 4</math>. <math>(a_1, a_2)</math> is a permutation of <math>(1, 2)</math>, <math>(1, 3)</math>, or <math>(2, 3)</math>, which gives a total of <math>3(2 \cdot 2) = 12</math> solutions.
 +
 
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Case 5: <math>a_3 = 5</math>. <math>(a_1, a_2)</math> is a permutation of <math>(1, 2)</math> or <math>(1, 3)</math>, which gives a total of <math>2(2 \cdot 2) = 8</math> solutions.
 +
 
 +
Therefore, our answer is <math>8 + 12 + 8 + 12 + 8 = 48 \Rightarrow \boxed{D}</math>.
 +
 
 +
-MP8148
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2008|num-b=20|num-a=22|ab=A}}
 
{{AMC12 box|year=2008|num-b=20|num-a=22|ab=A}}
  
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[[Category:Intermediate Geometry Problems]]
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[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 00:21, 31 July 2023

Problem

A permutation $(a_1,a_2,a_3,a_4,a_5)$ of $(1,2,3,4,5)$ is heavy-tailed if $a_1 + a_2 < a_4 + a_5$. What is the number of heavy-tailed permutations?

$\mathrm{(A)}\ 36\qquad\mathrm{(B)}\ 40\qquad\textbf{(C)}\ 44\qquad\mathrm{(D)}\ 48\qquad\mathrm{(E)}\ 52$

Solution 1 (Symmetry)

There are $5!=120$ total permutations.

For every permutation $(a_1,a_2,a_3,a_4,a_5)$ such that $a_1 + a_2 < a_4 + a_5$, there is exactly one permutation such that $a_1 + a_2 > a_4 + a_5$. Thus it suffices to count the permutations such that $a_1 + a_2 = a_4 + a_5$.

$1+4=2+3$, $1+5=2+4$, and $2+5=3+4$ are the only combinations of numbers that can satisfy $a_1 + a_2 = a_4 + a_5$.

There are $3$ combinations of numbers, $2$ possibilities of which side of the equation is $a_1+a_2$ and which side is $a_4+a_5$, and $2^2=4$ possibilities for rearranging $a_1,a_2$ and $a_4,a_5$. Thus, there are $3\cdot2\cdot4=24$ permutations such that $a_1 + a_2 = a_4 + a_5$.

Thus, the number of heavy-tailed permutations is $\frac{120-24}{2}=48 \Rightarrow D$.

Solution 2 (Casework)

We use case work on the value of $a_3$.

Case 1: $a_3 = 1$. Since $a_1 + a_2 < a_4 + a_5$, $(a_1, a_2)$ can only be a permutation of $(2, 3)$ or $(2, 4)$. The values of $a_1$ and $a_2$, as well as the values of $a_4$ and $a_5$, are interchangeable, so this case produces a total of $2(2 \cdot 2) = 8$ solutions.

Case 2: $a_3 = 2$. Similarly, we have $(a_1, a_2)$ is a permutation of $(1, 3)$, $(1, 4)$, or $(1, 5)$, which gives a total of $3(2 \cdot 2) = 12$ solutions.

Case 3: $a_3 = 3$. $(a_1, a_2)$ is a permutation of $(1, 2)$ or $(1, 4)$, which gives a total of $2(2 \cdot 2) = 8$ solutions.

Case 4: $a_3 = 4$. $(a_1, a_2)$ is a permutation of $(1, 2)$, $(1, 3)$, or $(2, 3)$, which gives a total of $3(2 \cdot 2) = 12$ solutions.

Case 5: $a_3 = 5$. $(a_1, a_2)$ is a permutation of $(1, 2)$ or $(1, 3)$, which gives a total of $2(2 \cdot 2) = 8$ solutions.

Therefore, our answer is $8 + 12 + 8 + 12 + 8 = 48 \Rightarrow \boxed{D}$.

-MP8148

See also

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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