1991 AIME Problems/Problem 2
Contents
Problem
Rectangle has sides
of length 4 and
of length 3. Divide
into 168 congruent segments with points
, and divide
into 168 congruent segments with points
. For
, draw the segments
. Repeat this construction on the sides
and
, and then draw the diagonal
. 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]](http://latex.artofproblemsolving.com/5/5/7/5578d04cdc2b7f7fa139b6bb512b5125dd621ed5.png)
The length of the diagonal is (a 3-4-5 right triangle). For each
,
is the hypotenuse of a
right triangle with sides of
. Thus, its length is
. Let
. We want to find
since we are over counting the diagonal.
Solution 2
Using the above diagram, we have that and each one of these is a dilated 3-4-5 right triangle (This is true since
is a 3-4-5 right triangle). Now, for all
, we have that
is the hypotenuse for the triangle
. Therefore we want to know the sum of the lengths of all
.This is given by the following:
Then by the summation formula for the sum of the terms of an arithmetic series,
~qwertysri987
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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