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1999 AHSME Problems/Problem 19

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Problem

Consider all triangles ABC satisfying in the following conditions: AB = AC, D is a point on \overline{AC} for which \overline{BD} \perp \overline{AC}, AC and CD are integers, and BD^{2} = 57. Among all such triangles, the smallest possible value of AC is

pair A,B,C,D; A=(5,12); B=origin; C=(10,0); D=(8.52071005917,3.55029585799);draw(A--B--C--cycle); draw(B--D);label("$A$&...

\textrm{(A)} \ 9 \qquad \textrm{(B)} \ 10 \qquad \textrm{(C)} \ 11 \qquad \textrm{(D)} \ 12 \qquad \textrm{(E)} \ 13

Solution

Thus AD = AC - CD and AB = AC are integers. By the Pythagorean Theorem,

AD^2 + 57 = AB^2 \Longrightarrow 1 \cdot 57 = 3 \cdot 19 = (AB - AD)(AB + AD).

Thus AC = AB = \frac {1 + 57}{2} = 29 or \frac {3 + 19}{2} = 11 \Longrightarrow \mathrm{(C)}.

See also

1999 AHSME (Problems)
Preceded by
Problem 18
Followed by
Problem 20
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