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

Revision as of 22:37, 19 February 2008

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?

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

Solution

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 $\frac{120-24}{2}=48 \Rightarrow D$ .

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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All AMC 12 Problems and Solutions

@@ Line 1: / Line 1: @@
-== Problem ==
+==Problem==
-[[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>?
+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?
-<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) \\
+<math>\textbf{(A)}\ 36 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 44 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 52</math>
-\textbf{(E)}\ \frac {3}{28}\left(10 - \sqrt {2}\right)</math>
-== Solution ==
+==Solution==
-<center><asy>
+There are <math>5!=120</math> total permutations.
-import olympiad;
-size(300);
-defaultpen(0.8);
-pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429);
-pair O=incenter(A,C,D), P=incenter(B,C,D);
-picture p = new picture;
-draw(p,Circle(C,0.2)); draw(p,Circle(B,0.2));
-clip(p,B--C--D--cycle);
-add(p);
-draw(A--B--C--D--C--cycle);
-draw(incircle(A,C,D));
-draw(incircle(B,C,D));
-dot(O);dot(P);
-label("\(A\)",A,W);
-label("\(B\)",B,E);
-label("\(C\)",C,W);
-label("\(D\)",D,NE);
-label("\(O_A\)",O,W);
-label("\(O_B\)",P,W);
-label("\(3\)",(A+C)/2,W);
-label("\(4\)",(B+C)/2,S);
-label("\(\frac{15}{7}\)",(A+D)/2,NE);
-label("\(\frac{20}{7}\)",(B+D)/2,NE);
-label("\(45^{\circ}\)",(.2,.1),E);
-label("\(\sin \theta = \frac{3}{5}\)",B-(.2,-.1),W);
-</asy></center>
-By the [[Angle Bisector Theorem]],
+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>.
-<cmath>\frac{BD}{4} = \frac{5-BD}{3} \Longrightarrow BD = \frac{20}7</cmath>
-By [[Law of Sines]] on <math>\triangle BCD</math>,
+<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>.
-<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>
-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
+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>.
-<cmath>\frac{r_A}{r_B} = \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A}</cmath>
-<!--Using any of various formulas for triangle area, we find the area <math>[BCD]</math> to be
+Thus, the number of <u>heavy-tailed</u> permutations is <math>\frac{120-24}{2}=48 \Rightarrow D</math>.
-<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>
-and
-<cmath>[ACD] = [ABC] - [BCD] = \frac 12 (3)(4) - \frac{24}{7} = \frac{18}{7}</cmath>-->
-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>
-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,
-<cmath>\begin{align*}
-\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} \\
-&= \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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Difference between revisions of "2008 AMC 12A Problems/Problem 21"

Revision as of 22:37, 19 February 2008

Problem

Solution

See also