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Difference between revisions of "1991 AIME Problems/Problem 2"

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</asy></center><!-- asy replacing Image:1991 AIME-2.png by azjps -->
 
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The length of the diagonal is <math>\sqrt{3^2 + 4^2} = 5</math> (a 3-4-5 [[right triangle]]). For each <math>k</math>, <math>\overline{P_kQ_k}</math> is the [[hypotenuse]] of a <math>3-4-5</math> right triangle with sides of <math>3 \cdot \frac{k}{168}, 4 \cdot \frac{k}{168}</math>. Thus, its length is <math>5 \cdot \frac{k}{168}</math>. Let <math>a_k=\frac{5k}{168}</math>. We want to find 2<math>\sum\limits_{k=1}^{168} a_k-5</math> since we are over counting the diagonal.
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The length of the diagonal is <math>\sqrt{3^2 + 4^2} = 5</math> (a 3-4-5 [[right triangle]]). For each <math>k</math>, <math>\overline{P_kQ_k}</math> is the [[hypotenuse]] of a <math>3-4-5</math> right triangle with sides of <math>3 \cdot \frac{k}{168}, 4 \cdot \frac{k}{168}</math>. Thus, its length is <math>5 \cdot \frac{k}{168}</math>. Let <math>a_k=\frac{5k}{168}</math>. We want to find <math>2\sum\limits_{k=1}^{168} a_k-5</math> since we are over counting the diagonal.
 
<math>2\sum\limits_{k=1}^{168} \frac{5k}{168}-5
 
<math>2\sum\limits_{k=1}^{168} \frac{5k}{168}-5
 
=2\sum\limits_{k=0}^{168} \frac{5k}{168}-5
 
=2\sum\limits_{k=0}^{168} \frac{5k}{168}-5

Revision as of 23:49, 2 January 2016

Problem

Rectangle $ABCD_{}^{}$ has sides $\overline {AB}$ of length 4 and $\overline {CB}$ of length 3. Divide $\overline {AB}$ into 168 congruent segments with points $A_{}^{}=P_0, P_1, \ldots, P_{168}=B$, and divide $\overline {CB}$ into 168 congruent segments with points $C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B$. For $1_{}^{} \le k \le 167$, draw the segments $\overline {P_kQ_k}$. Repeat this construction on the sides $\overline {AD}$ and $\overline {CD}$, and then draw the diagonal $\overline {AC}$. Find the sum of the lengths of the 335 parallel segments drawn.

Solution

[asy] real r = 0.35; size(220); pointpen=black;pathpen=black+linewidth(0.65);pen f = fontsize(8); pair A=(0,0),B=(4,0),C=(4,3),D=(0,3); D(A--B--C--D--cycle); pair P1=A+(r,0),P2=A+(2r,0),P3=B-(r,0),P4=B-(2r,0); pair Q1=C-(0,r),Q2=C-(0,2r),Q3=B+(0,r),Q4=B+(0,2r); D(A--C);D(P1--Q1);D(P2--Q2);D(P3--Q3);D(P4--Q4); MP("A",A,f);MP("B",B,SE,f);MP("C",C,NE,f);MP("D",D,W,f); MP("P_1",P1,f);MP("P_2",P2,f);MP("P_{167}",P3,f);MP("P_{166}",P4,f);MP("Q_1",Q1,E,f);MP("Q_2",Q2,E,f);MP("Q_{167}",Q3,E,f);MP("Q_{166}",Q4,E,f); MP("4",(A+B)/2,N,f);MP("\cdots",(A+B)/2,f); MP("3",(B+C)/2,W,f);MP("\vdots",(C+B)/2,E,f); [/asy]

The length of the diagonal is $\sqrt{3^2 + 4^2} = 5$ (a 3-4-5 right triangle). For each $k$, $\overline{P_kQ_k}$ is the hypotenuse of a $3-4-5$ right triangle with sides of $3 \cdot \frac{k}{168}, 4 \cdot \frac{k}{168}$. Thus, its length is $5 \cdot \frac{k}{168}$. Let $a_k=\frac{5k}{168}$. We want to find $2\sum\limits_{k=1}^{168} a_k-5$ since we are over counting the diagonal. $2\sum\limits_{k=1}^{168} \frac{5k}{168}-5 =2\sum\limits_{k=0}^{168} \frac{5k}{168}-5 =2\frac{(0+5)\cdot169}{2}-5 =168\cdot5 =\boxed{840}$

See also

1991 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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