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

(Created page with "== Problem == Diameter <math> \overline{AB}</math> of a circle with center <math> O</math> is <math> 10</math> units. <math> C</math> is a point <math> 4</math> units from <math>...")
 
(→‎Solution)
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== Solution ==
 
== Solution ==
 
<math>\fbox{}</math>
 
<math>\fbox{}</math>
 +
wait i need to solve it first
  
 
== See Also ==
 
== See Also ==

Revision as of 23:02, 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

$\fbox{}$ wait i need to solve it first

See Also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 47 		Followed by
Problem 49
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All AHSME Problems and Solutions

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