Problem:
Triangle ABC is an isosceles right triangle with AB=AC=3. Let M be the midpoint of hypotenuse BC. Points I and E lie on sides AC and AB, respectively, so that AI>AE and AIME is a cyclic quadrilateral. Given that triangle EMI has area 2, the length CI can be written as ca−b, where a,b, and c are positive integers and b is not divisible by the square of any prime. What is the value of a+b+c?
Answer Choices:
A. 9
B. 10
C. 11
D. 12
E. 13 Solution:
It follows from the Pythagorean Theorem that CM=MB=232. Because quadrilateral AIME is cyclic, opposite angles are supplementary and thus ∠IME is a right angle. Let x=CI and y=BE; then AI=3−x and AE=3−y. By the Law of Cosines in △MCI,
IM2=x2+(232)2−2⋅x⋅232⋅cos45∘=x2−3x+29
Similarly, ME2=y2−3y+29. By the Pythagorean Theorem in right triangles EMI and IAE,
(x2−3x+29)+(y2−3y+29)=(3−x)2+(3−y)2
which simplifies to x+y=3. Because the area of △EMI is 2 , it follows that IM2⋅ME2=16. Therefore
(x2−3x+29)((3−x)2−3(3−x)+29)=16
which simplifies to (x2−3x+29)2=16. Because y>x, the only real solution is x=23−7. The requested sum is 3+7+2=12.
OR
Place the figure in the coordinate plane with A at (0,0),B at (3,0), and C at (0,3). Then M is at (23,23). Let s=AE and t=CI. Then the coordinates of E are (s,0), and the coordinates of I are (0,3−t). Because AIME is a cyclic quadrilateral and ∠EAI is a right angle, ∠IME is a right angle. Therefore MI and ME are perpendicular, so the product of their slopes is
23−s23⋅23t−23=−1
this equation simplifies to s=t. Then, with brackets indicating area,
