Difference between revisions of "1958 AHSME Problems/Problem 48"
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== Solution == | == Solution == | ||
<math>\fbox{}</math> | <math>\fbox{}</math> | ||
| − | + | *If somebody wants to draw a diagram or make this solution better, PLEASE do so. I cannot express how bad I am at this. | |
| + | If P is on A, then the length is 10, eliminating answer choice (B). | ||
| + | If P is equidistant from C and D, the length is 2\sqrt{1^{2}+5^{2}}=2\sqrt{26}>10, eliminating (A) and (C). | ||
| + | If CDP is a right triangle, then CDP will be right or DCP will be right. Assume that DCP is right. | ||
== See Also == | == See Also == | ||
Revision as of 23:15, 31 December 2023
Problem
Diameter 
of a circle with center 
is 
units. 
is a point 
units from 
, and on . 
is a point units from 
, and on . 
is any point on the circle. Then the broken-line path from to to :
Solution
- If somebody wants to draw a diagram or make this solution better, PLEASE do so. I cannot express how bad I am at this.
If P is on A, then the length is 10, eliminating answer choice (B). If P is equidistant from C and D, the length is 2\sqrt{1^{2}+5^{2}}=2\sqrt{26}>10, eliminating (A) and (C). If CDP is a right triangle, then CDP will be right or DCP will be right. Assume that DCP is right.
See Also
| 1958 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 47 |
Followed by Problem 49 | |
| 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. 
