Difference between revisions of "1991 AIME Problems/Problem 2"
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D(A--C);D(P1--Q1);D(P2--Q2);D(P3--Q3);D(P4--Q4); | 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("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_{ | + | 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("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); | MP("3",(B+C)/2,W,f);MP("\vdots",(C+B)/2,E,f); |
Revision as of 20:15, 27 August 2014
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
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 .
The sum we are looking for is . Using the formula for the sum of the first natural numbers, we find that the solution is .
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.