Question

Triangle $ABC$ is an isosceles right triangle with $AB = AC = 3.$ Let $M$ be the midpoint of hypotenuse $\overline{BC}.$ Points $I$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB},$ respectively, so that $AI \gt AE$ and $AIME$ is a cyclic quadrilateral. Given that triangle $EMI$ has area $2,$ the length $CI$ can be written as $\tfrac{a - \sqrt{b}}{c},$ where $a,$ $b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is the value of $a + b + c?$

$9$

$10$

$11$

$12$

$13$

Answer: D

Concepts:cyclic quadrilateral law of cosines triangle area

Difficulty rating: 2110

Solution:

Since $\triangle ABC$ is an isosceles right triangle, $CM = BM = \tfrac32 \sqrt2$ and the base angles at $B, C$ are $45^\circ.$ As $AIME$ is cyclic with right angle at $A,$ angle $\angle IME = 90^\circ.$ Let $x = CI$ and $y = BE.$ By the Law of Cosines in $\triangle MCI,$ $\begin{aligned} IM^2 &= x^2 + \tfrac92 \\ &\quad {}- 2 \cdot x \cdot \tfrac32\sqrt2 \cdot \cos 45^\circ \\ &= x^2 - 3x + \tfrac92, \end{aligned}$ and similarly $ME^2 = y^2 - 3y + \tfrac92.$

The Pythagorean Theorem in right triangles $EMI$ and $IAE$ gives $IM^2 + ME^2$ $= (3-x)^2$ $+ (3-y)^2,$ which simplifies to $x + y = 3.$ The area condition $\tfrac12 IM \cdot ME = 2$ means $IM^2 \cdot ME^2 = 16.$ Substituting $y = 3 - x$ makes $ME^2 = x^2 - 3x + \tfrac92 = IM^2,$ so $\left(x^2 - 3x + \tfrac92\right)^2 = 16,$ hence $x^2 - 3x + \tfrac92 = 4,$ i.e. $x^2 - 3x + \tfrac12 = 0.$

Since $AI \gt AE$ forces $y \gt x,$ we take the smaller root $x = \tfrac{3 - \sqrt{7}}{2}.$ Then $a + b + c = 3 + 7 + 2 = 12.$

Thus, the correct answer is D.

2018 AMC 12A Problem 20

Answer: D

Difficulty rating: 2110

Solution:

Problem 20 in Other Years