Difference between revisions of "1991 AIME Problems/Problem 2"
(→Solution) |
m (→Solution) |
||
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{ | + | 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{ | + | <math>2\sum\limits_{k=1}^{168} \frac{5(168-k)}{168}-5 |
− | |||
=2\frac{(0+5)\cdot169}{2}-5 | =2\frac{(0+5)\cdot169}{2}-5 | ||
=168\cdot5 | =168\cdot5 |
Revision as of 22:48, 21 May 2019
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
![[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.
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.