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

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== Solution ==
 
== Solution ==
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*If somebody wants to draw a diagram or make this solution better, PLEASE do so. I cannot express how bad I am at this.
 
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If P is on A, then the length is 10, eliminating answer choice (B).
 
  
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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).
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If <math>P</math> is on <math>A</math>, then the length is 10, eliminating answer choice <math>(B)</math>.
  
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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).
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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>.
  
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Thus, our answer is (E).
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If triangle <math>CDP</math> is right, then angle <math>CDP</math> is right or angle <math>DCP</math> is right. Assume that angle <math>DCP</math> is right. Triangle <math>APB</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 <math>(D)</math>.
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Thus, our answer is <math>(E)</math>.
 
<math>\fbox{}</math>
 
<math>\fbox{}</math>
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Note: Say you are not convinced that <math>\sqrt{24}+\sqrt{28}>10</math>. We can prove this as follows.
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Start by simplifying the equation: <math>\sqrt{6}+\sqrt{7}>5</math>.
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Square both sides: <math>6+2\sqrt{42}+7>25</math>.
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Simplify: <math>\sqrt{42}>6</math>
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Square both sides again: <math>42>36</math>. From here, we can just reverse our steps to get <math>\sqrt{24}+\sqrt{28}>10</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 23:49, 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 $P$ 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 triangle $CDP$ is right, then angle $CDP$ is right or angle $DCP$ is right. Assume that angle $DCP$ is right. Triangle $APB$ 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 $(E)$. $\fbox{}$

Note: Say you are not convinced that $\sqrt{24}+\sqrt{28}>10$. We can prove this as follows.

Start by simplifying the equation: $\sqrt{6}+\sqrt{7}>5$.

Square both sides: $6+2\sqrt{42}+7>25$.

Simplify: $\sqrt{42}>6$

Square both sides again: $42>36$. From here, we can just reverse our steps to get $\sqrt{24}+\sqrt{28}>10$.

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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