Difference between revisions of "1951 AHSME Problems/Problem 32"
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If <math>\triangle ABC</math> is inscribed in a semicircle whose diameter is <math>AB</math>, then <math>AC+BC</math> must be | If <math>\triangle ABC</math> is inscribed in a semicircle whose diameter is <math>AB</math>, then <math>AC+BC</math> must be | ||
| − | <math> \textbf{(A)}\ \text{equal to }AB\qquad\textbf{(B)}\ \text{equal to }AB\sqrt{2}\qquad\textbf{(C)}\ \geq AB\sqrt{2}\qquad\textbf{(D)}\ \leq AB\sqrt{2} | + | <math> \textbf{(A)}\ \text{equal to }AB\qquad\textbf{(B)}\ \text{equal to }AB\sqrt{2}\qquad\textbf{(C)}\ \geq AB\sqrt{2}\qquad\textbf{(D)}\ \leq AB\sqrt{2} \qquad \textbf{(E)}\ AB^{2} </math> |
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==Solution== | ==Solution== | ||
Latest revision as of 20:10, 21 June 2024
Problem
If 
is inscribed in a semicircle whose diameter is 
, then 
must be
Solution
Because is the diameter of the semi-circle, it follows that 
. Now we can try to eliminate all the solutions except for one by giving counterexamples.
Set point 
anywhere on the perimeter of the semicircle except on . By triangle inequality, 
, so 
is wrong.
Set point on the perimeter of the semicircle infinitesimally close to , and so almost equals , therefore 
is wrong.
Because we proved that can be very close to in case , it follows that 
is wrong.
Because we proved that can be very close to in case , it follows that 
is wrong.
Therefore, the only possible case is 
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 31 |
Followed by Problem 33 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
