Difference between revisions of "1992 AHSME Problems/Problem 20"
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== Solution == | == Solution == | ||
− | If we sum up the angles to obtain 360, we can see that the B angles add to the sum and the A angles subtract from the sum (an easy way of looking at this is by using the opposing angle theorem: if A[n] = B[n] than their total contribution is 0). Thus we have B[1] + B[2] + ... + B[n] - A[1] - A[2] - ... A[n] = 360. But every pair of A,B has a total 'angle contribution' of 10, thus there are 36 <math>\fbox{D}</math> pairs of A, | + | If we sum up the angles to obtain 360, we can see that the B angles add to the sum and the A angles subtract from the sum (an easy way of looking at this is by using the opposing angle theorem: if A[n] = B[n] than their total contribution is 0). Thus we have B[1] + B[2] + ... + B[n] - A[1] - A[2] - ... A[n] = 360. But every pair of A,B has a total 'angle contribution' of 10, thus there are 36 <math>\fbox{D}</math> pairs of A,B. |
== See also == | == See also == |
Latest revision as of 12:28, 1 May 2021
Problem
Part of an "n-pointed regular star" is shown. It is a simple closed polygon in which all
edges are congruent, angles
are congruent, and angles
are congruent. If the acute angle at
is
less than the acute angle at
, then
Solution
If we sum up the angles to obtain 360, we can see that the B angles add to the sum and the A angles subtract from the sum (an easy way of looking at this is by using the opposing angle theorem: if A[n] = B[n] than their total contribution is 0). Thus we have B[1] + B[2] + ... + B[n] - A[1] - A[2] - ... A[n] = 360. But every pair of A,B has a total 'angle contribution' of 10, thus there are 36 pairs of A,B.
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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