Difference between revisions of "1999 AHSME Problems/Problem 19"
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Consider all [[triangle]]s <math>ABC</math> satisfying in the following conditions: <math>AB = AC</math>, <math>D</math> is a point on <math>\overline{AC}</math> for which <math>\overline{BD} \perp \overline{AC}</math>, <math>AC</math> and <math>CD</math> are integers, and <math>BD^{2} = 57</math>. Among all such triangles, the smallest possible value of <math>AC</math> is | Consider all [[triangle]]s <math>ABC</math> satisfying in the following conditions: <math>AB = AC</math>, <math>D</math> is a point on <math>\overline{AC}</math> for which <math>\overline{BD} \perp \overline{AC}</math>, <math>AC</math> and <math>CD</math> are integers, and <math>BD^{2} = 57</math>. Among all such triangles, the smallest possible value of <math>AC</math> 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> | ||
<math>\textrm{(A)} \ 9 \qquad \textrm{(B)} \ 10 \qquad \textrm{(C)} \ 11 \qquad \textrm{(D)} \ 12 \qquad \textrm{(E)} \ 13</math> | <math>\textrm{(A)} \ 9 \qquad \textrm{(B)} \ 10 \qquad \textrm{(C)} \ 11 \qquad \textrm{(D)} \ 12 \qquad \textrm{(E)} \ 13</math> | ||
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:35, 5 July 2013
Problem
Consider all triangles 
satisfying in the following conditions: 
, 
is a point on 
for which 
, 
and 
are integers, and 
. Among all such triangles, the smallest possible value of 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]](http://latex.artofproblemsolving.com/4/d/0/4d08273e2a3ddb2c8fffd607582fa89298ab91db.png)
Solution
Thus 
and are integers. By the Pythagorean Theorem,
![\[AD^2 + 57 = AB^2 \Longrightarrow 1 \cdot 57 = 3 \cdot 19 = (AB - AD)(AB + AD).\]](http://latex.artofproblemsolving.com/3/b/a/3ba85b5e9b5d2e89a45a9e4d3843ac17664a2818.png)
Thus 
or 
.
See also
| 1999 AHSME (Problems • Answer Key • Resources) | ||
| 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. 
