Difference between revisions of "1958 AHSME Problems/Problem 42"

Revision as of 11:00, 27 November 2021

Problem

In a circle with center $O$ , chord $\overline{AB}$ equals chord $\overline{AC}$ . Chord $\overline{AD}$ cuts $\overline{BC}$ in $E$ . If $AC = 12$ and $AE = 8$ , then $AD$ equals:

$\textbf{(A)}\ 27\qquad \textbf{(B)}\ 24\qquad \textbf{(C)}\ 21\qquad \textbf{(D)}\ 20\qquad \textbf{(E)}\ 18$

Solution

Let $X$ be a point on $BC$ so $AX \perp BC$ . Let $AX = h$ , $EX = \sqrt{64 - h^2}$ and $BX = \sqrt{144 - h^2}$ . $CE = CX - EX = \sqrt{144 - h^2} - \sqrt{64 - h^2}$ . Using Power of a Point on $E$ , $(BE)(EC) = (AE)(ED)$ (there isn't much information about the circle so I wanted to use PoP).

$(\sqrt{144 - h^2} - \sqrt{64 - h^2})(\sqrt{144 - h^2} + \sqrt{64 - h^2}) = 8(ED)$

$(144 - h^2) - (64 - h^2) = 8(ED)$

$80 = 8(ED)$

$ED = 10$

Adding up $AD$ and $ED$ we get $\fbox{E}$ .

1958 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 41	Followed by Problem 43
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 16: / Line 16: @@
 <cmath>(144 - h^2) - (64 - h^2) = 8(ED)</cmath>
-<math>80 = 8(ED)<cmath>
+<cmath>80 = 8(ED)</cmath>
-</cmath>ED = 10</math><math>
+<cmath>ED = 10</cmath>
-Adding up </math>AD<math> and </math>ED<math> we get </math>\fbox{E}$
+Adding up <math>AD</math> and <math>ED</math> we get <math>\fbox{E}</math>.
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Difference between revisions of "1958 AHSME Problems/Problem 42"

Revision as of 11:00, 27 November 2021

Problem

Solution

See Also