2015 AIME I Problems/Problem 11
Triangle has positive integer side lengths with . Let be the intersection of the bisectors of and . Suppose . Find the smallest possible perimeter of .
Let be the midpoint of . Then by SAS Congruence, , so .
Now let , , and .
Cross-multiplying yields .
Since , must be positive, so .
Additionally, since has hypotenuse of length , .
Therefore, given that is an integer, the only possible values for are , , , and .
However, only one of these values, , yields an integral value for , so we conclude that and .
Thus the perimeter of must be .
Solution 2 (No Trig)
Let and the foot of the altitude from to be point and . Since ABC is isosceles, is on . By Pythagorean Theorem, . Let and . By Angle Bisector theorem, . Also, . Solving for , we get . Then, using Pythagorean Theorem on we have . Simplifying, we have . Factoring out the , we have . Adding 1 to the fraction and simplifying, we have . Crossing out the , and solving for yields . Then, we continue as Solution 1 does.
Let , call the midpoint of point , call the point where the incircle meets point ,
and let . We are looking for the minimum value of . is an altitude because the triangle
is isosceles. By Pythagoras on , the inradius is and by Pythagoras on , is
. By equal tangents, , so . Since is an inradius, and using pythagoras on yields . is similar to by , so we
can write . Simplifying, .
Squaring, subtracting 1 from both sides, and multiplying everything out, we get , which turns into . Finish as in Solution 1.
Angle bisectors motivate trig bash. Define angle . Foot of perpendicular from to is point . , where is an integer. Thus, . Via double angle, we calculate to be . This is to be an integer. We can bound now, as to avoid negative values and due to triangle inequality. Testing, works, giving . Our answer is . - whatRthose
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