Difference between revisions of "1958 AHSME Problems/Problem 48"
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== Solution == | == Solution == | ||
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− | If P is on A, then the length is 10, eliminating answer choice (B). | + | 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 | + | |
− | If CDP is | + | 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 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>. | ||
+ | |||
+ | Thus, our answer is <math>(E)</math>. | ||
<math>\fbox{}</math> | <math>\fbox{}</math> | ||
+ | |||
+ | Note: Say you are not convinced that <math>\sqrt{24}+\sqrt{28}>10</math>. We can prove this as follows. | ||
+ | |||
+ | Start by simplifying the equation: <math>\sqrt{6}+\sqrt{7}>5</math>. | ||
+ | |||
+ | Square both sides: <math>6+2\sqrt{42}+7>25</math>. | ||
+ | |||
+ | Simplify: <math>\sqrt{42}>6</math> | ||
+ | |||
+ | 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 00:49, 1 January 2024
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 is on
, then the length is 10, eliminating answer choice
.
If is equidistant from
and
, the length is
, eliminating
and
.
If triangle is right, then angle
is right or angle
is right. Assume that angle
is right. Triangle
is right, so
. Then,
, so the length we are looking for is
, eliminating
.
Thus, our answer is .
Note: Say you are not convinced that . We can prove this as follows.
Start by simplifying the equation: .
Square both sides: .
Simplify:
Square both sides again: . From here, we can just reverse our steps to get
.
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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