Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(Adding another possible solution)
 
(5 intermediate revisions by 3 users not shown)
Line 2: Line 2:
 
[[Rectangle]] <math>ABCD_{}^{}</math> has [[edge | sides]] <math>\overline {AB}</math> of [[length]] 4 and <math>\overline {CB}</math> of length 3. Divide <math>\overline {AB}</math> into 168 [[congruent]] [[segment]]s with [[point]]s <math>A_{}^{}=P_0, P_1, \ldots, P_{168}=B</math>, and divide <math>\overline {CB}</math> into 168 congruent segments with points <math>C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B</math>. For <math>1_{}^{} \le k \le 167</math>, draw the segments <math>\overline {P_kQ_k}</math>. Repeat this [[construction]] on the sides <math>\overline {AD}</math> and <math>\overline {CD}</math>, and then draw the [[diagonal]] <math>\overline {AC}</math>. Find the sum of the lengths of the 335 [[parallel]] segments drawn.
 
[[Rectangle]] <math>ABCD_{}^{}</math> has [[edge | sides]] <math>\overline {AB}</math> of [[length]] 4 and <math>\overline {CB}</math> of length 3. Divide <math>\overline {AB}</math> into 168 [[congruent]] [[segment]]s with [[point]]s <math>A_{}^{}=P_0, P_1, \ldots, P_{168}=B</math>, and divide <math>\overline {CB}</math> into 168 congruent segments with points <math>C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B</math>. For <math>1_{}^{} \le k \le 167</math>, draw the segments <math>\overline {P_kQ_k}</math>. Repeat this [[construction]] on the sides <math>\overline {AD}</math> and <math>\overline {CD}</math>, and then draw the [[diagonal]] <math>\overline {AC}</math>. Find the sum of the lengths of the 335 [[parallel]] segments drawn.
  
−
== Solution ==
+
== Solution 1 ==
 
<center><asy>
 
<center><asy>
 
real r = 0.35; size(220);
 
real r = 0.35; size(220);
Line 17: Line 17:
 
</asy></center><!-- asy replacing Image:1991 AIME-2.png by azjps -->
 
</asy></center><!-- asy replacing Image:1991 AIME-2.png by azjps -->
  
−
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</math>-5 since we are over counting <math>5</math>.
+
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{168-k}{168}, 4 \cdot \frac{168-k}{168}</math>. Thus, its length is <math>5 \cdot \frac{168-k}{168}</math>. Let <math>a_k=\frac{5(168-k)}{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{5(168-k)}{168}-5
−
=2\sum\limits_{k=0}^{168} \frac{5k}{168}-5
 
 
=2\frac{(0+5)\cdot169}{2}-5
 
=2\frac{(0+5)\cdot169}{2}-5
 
=168\cdot5
 
=168\cdot5
 
=\boxed{840}</math>
 
=\boxed{840}</math>
 +
 +
==Solution 2==
 +
 +
Using the above diagram, we have that <math> \Delta ABC \sim \Delta P_k B Q_k </math> and each one of these is a dilated 3-4-5 right triangle (This is true since <math>\Delta ABC </math> is a 3-4-5 right triangle). Now, for all <math>k</math>, we have that <math>\overline{P_k Q_k}</math> is the hypotenuse for the triangle <math>P_k B Q_k</math>. Therefore we want to know the sum of the lengths of all <math>\overline{P_k Q_k}</math>.This is given by the following:
 +
<cmath>  2 \cdot(\sum_{k=1}^{168} P_kQ_k) + 5 </cmath>
 +
 +
<cmath> = 2 \cdot \frac{ 0+5+10+...+835}{168} +5</cmath>
 +
Then by the summation formula for the sum of the terms of an arithmetic series,
 +
<cmath> = \frac{835 \cdot 168}{168} +5 = 835+5 = \boxed{840}</cmath>
 +
 +
~qwertysri987
 +
 +
==Solution 3==
 +
First, count the diagonal which has length <math>5</math>. For the rest of the segments, think about pairing them up so that each pair makes <math>5</math>. For example, the parallel lines closest to the diagonal would have length <math>\frac{167}{168}\cdot{5}</math> while the parallel line closest to the corner of the rectangle would have length <math>\frac{1}{168}\cdot{5}</math> by similar triangles. If you add the two lengths together, it is <math>\frac{167}{168}\cdot{5} + \frac{1}{168}\cdot{5} = 5.</math> There are <math>\frac{335-1}{2}</math> pairs of these segments, for a total of <math>5+(167)(5)=168(5)=\boxed{840}.</math>
 +
~justlearningmathog
  
 
== See also ==
 
== See also ==

Latest revision as of 11:51, 17 June 2021

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 1

[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 $3 \cdot \frac{168-k}{168}, 4 \cdot \frac{168-k}{168}$. Thus, its length is $5 \cdot \frac{168-k}{168}$. Let $a_k=\frac{5(168-k)}{168}$. We want to find $2\sum\limits_{k=1}^{168} a_k-5$ since we are over counting the diagonal. $2\sum\limits_{k=1}^{168} \frac{5(168-k)}{168}-5 =2\frac{(0+5)\cdot169}{2}-5 =168\cdot5 =\boxed{840}$

Solution 2

Using the above diagram, we have that $\Delta ABC \sim \Delta P_k B Q_k$ and each one of these is a dilated 3-4-5 right triangle (This is true since $\Delta ABC$ is a 3-4-5 right triangle). Now, for all $k$, we have that $\overline{P_k Q_k}$ is the hypotenuse for the triangle $P_k B Q_k$. Therefore we want to know the sum of the lengths of all $\overline{P_k Q_k}$.This is given by the following: \[2 \cdot(\sum_{k=1}^{168} P_kQ_k) + 5\]

\[= 2 \cdot \frac{ 0+5+10+...+835}{168} +5\] Then by the summation formula for the sum of the terms of an arithmetic series, \[= \frac{835 \cdot 168}{168} +5 = 835+5 = \boxed{840}\]

~qwertysri987

Solution 3

First, count the diagonal which has length $5$. For the rest of the segments, think about pairing them up so that each pair makes $5$. For example, the parallel lines closest to the diagonal would have length $\frac{167}{168}\cdot{5}$ while the parallel line closest to the corner of the rectangle would have length $\frac{1}{168}\cdot{5}$ by similar triangles. If you add the two lengths together, it is $\frac{167}{168}\cdot{5} + \frac{1}{168}\cdot{5} = 5.$ There are $\frac{335-1}{2}$ pairs of these segments, for a total of $5+(167)(5)=168(5)=\boxed{840}.$ ~justlearningmathog

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_2&oldid=156202"