2018 AMC 12A Problem 20

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

Small hint:

Because

AIME

is cyclic and

\angle A = 90^\circ,

the opposite angle

\angle IME = 90^\circ

as well

Big hint:

With

x = CI,

y = BE,

use the Law of Cosines for

IM, ME

(base angles

45^\circ

), the area

\tfrac12 IM \cdot ME = 2,

and the relation

x + y = 3

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.