Difference between revisions of "1958 AHSME Problems/Problem 42"
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== Problem == | == Problem == | ||
− | In a circle with center <math> O</math>, chord <math> \overline{AB}</math> equals chord <math> \overline{AC}</math>. Chord <math> \overline{AD}</math> cuts <math> \overline{BC}</math> in <math> E</math>. If <math> AC | + | In a circle with center <math> O</math>, chord <math> \overline{AB}</math> equals chord <math> \overline{AC}</math>. Chord <math> \overline{AD}</math> cuts <math> \overline{BC}</math> in <math> E</math>. If <math> AC = 12</math> and <math> AE = 8</math>, then <math> AD</math> equals: |
<math> \textbf{(A)}\ 27\qquad | <math> \textbf{(A)}\ 27\qquad | ||
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== Solution == | == Solution == | ||
− | <math>\fbox{}</math> | + | |
+ | Let <math>X</math> be a point on <math>BC</math> so <math>AX \perp BC</math>. Let <math>AX = h</math>, <math>EX = \sqrt{64 - h^2}</math> and <math>BX = \sqrt{144 - h^2}</math>. <math>CE = CX - EX = \sqrt{144 - h^2} - \sqrt{64 - h^2}</math>. Using Power of a Point on <math>E</math>, <math>(BE)(EC) = (AE)(ED)</math> (there isn't much information about the circle so I wanted to use PoP). | ||
+ | |||
+ | <cmath>(\sqrt{144 - h^2} - \sqrt{64 - h^2})(\sqrt{144 - h^2} + \sqrt{64 - h^2}) = 8(ED)</cmath> | ||
+ | |||
+ | <cmath>(144 - h^2) - (64 - h^2) = 8(ED)</cmath> | ||
+ | |||
+ | <cmath>80 = 8(ED)</cmath> | ||
+ | |||
+ | <cmath>ED = 10</cmath> | ||
+ | |||
+ | Adding up <math>AD</math> and <math>ED</math> we get <math>\fbox{E}</math>. | ||
== See Also == | == See Also == |
Latest revision as of 11:01, 29 June 2024
Problem
In a circle with center , chord
equals chord
. Chord
cuts
in
. If
and
, then
equals:
Solution
Let be a point on
so
. Let
,
and
.
. Using Power of a Point on
,
(there isn't much information about the circle so I wanted to use PoP).
Adding up and
we get
.
See Also
1958 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 41 |
Followed by Problem 43 | |
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