2018 AMC 12B Problem 21

Question

In

\triangle ABC

with side lengths

AB=13,

AC=12,

and

BC=5,

let

O

and

I

denote the circumcenter and incenter, respectively. A circle with center

M

is tangent to the legs

AC

and

BC

and to the circumcircle of

\triangle ABC.

What is the area of

\triangle MOI?

\dfrac{5}{2}

\dfrac{11}{4}

3

\dfrac{13}{4}

\dfrac{7}{2}

The triangle is right-angled at

C;

put

C

at the origin with the legs on the axes

O

is the midpoint of the hypotenuse and

I=(2,2);

the circle at

M=(\rho,\rho)

is tangent to the circumcircle

Since

5^2+12^2=13^2,

the triangle is right-angled at

C.

Set

C=(0,0),

A=(12,0),

and

B=(0,5).

Then

O

is the midpoint of

\overline{AB},

namely

O=\left(6,\tfrac52\right),

with circumradius

\tfrac{13}{2}.

The inradius is

\tfrac{\text{area}}{s}=\tfrac{30}{15}=2,

so

I=(2,2).

Because

M

’s circle is tangent to both legs,

M=(\rho,\rho).

Internal tangency to the circumcircle gives

MO=\tfrac{13}{2}-\rho.

Setting this equal to

\sqrt{(\rho-6)^2+\left(\rho-\tfrac52\right)^2}

and solving gives

\rho=4,

so

M=(4,4).

The shoelace formula on

M=(4,4),

O=\left(6,\tfrac52\right),

I=(2,2)

gives area

\tfrac72.

Thus, the correct answer is E.

Problem 21 in other years