Difference between revisions of "1981 AHSME Problems/Problem 25"
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Also, by Stewart’s Theorem, <math>198+11d^2=2b^2+9c^2</math> and <math>330+11e^2=5b^2+6c^2</math>. Therefore, we have the following system of equations using our substitution from earlier: | Also, by Stewart’s Theorem, <math>198+11d^2=2b^2+9c^2</math> and <math>330+11e^2=5b^2+6c^2</math>. Therefore, we have the following system of equations using our substitution from earlier: | ||
− | < | + | <cmath> |
Thus, we have: | Thus, we have: | ||
− | < | + | <cmath> |
Therefore, <math>5b^2=27c^2</math>, so <math>b^2=\frac{27c^2}5</math>, thus our first equation from earlier gives <math>264=\frac{33c^2}{5}</math>, so <math>c^2=40</math>, thus <math>b^2=216</math>. So, <math>c<b</math> and the answer to the original problem is <math>c=\sqrt{40}=\boxed{2\sqrt{10}~\textbf{(A)}}</math>. | Therefore, <math>5b^2=27c^2</math>, so <math>b^2=\frac{27c^2}5</math>, thus our first equation from earlier gives <math>264=\frac{33c^2}{5}</math>, so <math>c^2=40</math>, thus <math>b^2=216</math>. So, <math>c<b</math> and the answer to the original problem is <math>c=\sqrt{40}=\boxed{2\sqrt{10}~\textbf{(A)}}</math>. |
Revision as of 00:58, 9 August 2023
Problem 25
In in the adjoining figure, and trisect . The lengths of , and are , , and , respectively. The length of the shortest side of is
Solution
Let , , , and . Then, by the Angle Bisector Theorem, and , thus and .
Also, by Stewart’s Theorem, and . Therefore, we have the following system of equations using our substitution from earlier:
Thus, we have:
Therefore, , so , thus our first equation from earlier gives , so , thus . So, and the answer to the original problem is .
Aops-g5-gethsemanea2 (talk) 01:47, 9 August 2023 (EDT)
1981 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
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