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

(Solution)
(Solution)
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== Solution ==
 
== 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 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).
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If <math>P</math> is on <math>A</math>, then the length is 10, eliminating answer choice <math>(B)</math>.
  
If P is equidistant from C and D, the length is <math>2\sqrt{1^2+5^2}=2\sqrt{26}>10</math>, eliminating (A) and (C).
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If <math>P</math> is equidistant from <math>C</math> and <math>D</math>, the length is <math>2\sqrt{1^2+5^2}=2\sqrt{26}>10</math>, eliminating <math>(A)</math> and <math>(C)</math>.
  
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 <math>CP=\sqrt{4*6}=\sqrt{24}</math>. Then, <math>DP=\sqrt{28}</math>, so the length we are looking for is <math>\sqrt{24}+\sqrt{28}>10</math>, eliminating (D).
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If <math>\deltaCDP</math> is right, then <math>\angleCDP</math> is right or <math>\angleDCP</math> is right. Assume that <math>\angleDCP</math> is right. <math>\deltaAPB</math> is right, so <math>CP=\sqrt{4*6}=\sqrt{24}</math>. Then, <math>DP=\sqrt{28}</math>, so the length we are looking for is <math>\sqrt{24}+\sqrt{28}>10</math>, eliminating (D).
  
 
Thus, our answer is (E).
 
Thus, our answer is (E).

Revision as of 23:29, 31 December 2023

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 $\deltaCDP$ (Error compiling LaTeX. Unknown error_msg) is right, then $\angleCDP$ (Error compiling LaTeX. Unknown error_msg) is right or $\angleDCP$ (Error compiling LaTeX. Unknown error_msg) is right. Assume that $\angleDCP$ (Error compiling LaTeX. Unknown error_msg) is right. $\deltaAPB$ (Error compiling LaTeX. Unknown error_msg) 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 47
Followed by
Problem 49
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All AHSME Problems and Solutions

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