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Difference between revisions of "1958 AHSME Problems/Problem 48"

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If P is equidistant from C and D, the length is 2<math>\sqrt{1^2+5^2}</math>=2<math>\sqrt{26}</math>>10, eliminating (A) and (C).
 
If P is equidistant from C and D, the length is 2<math>\sqrt{1^2+5^2}</math>=2<math>\sqrt{26}</math>>10, eliminating (A) and (C).
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If CDP is a right triangle, then CDP will be right or DCP will be right. Assume that DCP is right. Then, APB is right, so CP=<math>\sqrt{4*6}</math>=<math>\sqrt{24}</math>. Then, DP=<math>\sqrt{28}</math>, so the length we are looking for is <math>\sqrt{24}</math>+<math>\sqrt{28}</math>>10, eliminating (D). Thus, our answer is (E).
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If CDP is a right triangle, then CDP will be right or DCP will be right. Assume that DCP is right. Then, APB is right, so CP=<math>\sqrt{4*6}</math>=<math>\sqrt{24}</math>. Then, DP=<math>\sqrt{28}</math>, so the length we are looking for is <math>\sqrt{24}</math>+<math>\sqrt{28}</math>>10, eliminating (D).
 +
 
 +
Thus, our answer is (E).
 
<math>\fbox{}</math>
 
<math>\fbox{}</math>
  

Revision as of 00:24, 1 January 2024

Problem

Diameter $\overline{AB}$ of a circle with center $O$ is $10$ units. $C$ is a point $4$ units from $A$, and on $\overline{AB}$. $D$ is a point $4$ units from $B$, and on $\overline{AB}$. $P$ is any point on the circle. Then the broken-line path from $C$ to $P$ to $D$:

$\textbf{(A)}\ \text{has the same length for all positions of }{P}\qquad\\ \textbf{(B)}\ \text{exceeds }{10}\text{ units for all positions of }{P}\qquad \\ \textbf{(C)}\ \text{cannot exceed }{10}\text{ units}\qquad \\ \textbf{(D)}\ \text{is shortest when }{\triangle CPD}\text{ is a right triangle}\qquad \\ \textbf{(E)}\ \text{is longest when }{P}\text{ is equidistant from }{C}\text{ and }{D}.$


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. Then, APB is right, so CP=$\sqrt{4*6}$=$\sqrt{24}$. Then, DP=$\sqrt{28}$, so the length we are looking for is $\sqrt{24}$+$\sqrt{28}$>10, eliminating (D).

Thus, our answer is (E). $\fbox{}$

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

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