During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

1999 AHSME Problems/Problem 19

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

[asy] 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("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); [/asy]

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

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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
All AHSME Problems and Solutions

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

Invalid username
Login to AoPS