2013 AIME II Problem 10

Attempt Problem 10 of the 2013 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AIME II solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

10.

Given a circle of radius

\sqrt{13},

let

A

be a point at a distance

4 + \sqrt{13}

from the center

O

of the circle. Let

B

be the point on the circle nearest to point

A.

A line passing through the point

A

intersects the circle at points

K

and

L.

The maximum possible area for

\triangle BKL

can be written in the form

\frac{a - b\sqrt{c}}{d},

where

a,

b,

c,

and

d

are positive integers,

a

and

d

are relatively prime, and

c

is not divisible by the square of any prime. Find

a + b + c + d.

Answer

Answer: 146

Concepts:circle triangle area optimization

Difficulty rating: 2840

Small hint:

B

lies on segment

OA

with

AB = 4.

For any line through

A,

the distances from

O

and from

B

to that line are in the ratio

OA : AB

Big hint:

So

[BKL] = \frac{4}{4 + \sqrt{13}}[OKL],

and

[OKL] = \frac{13}{2}\sin(\angle KOL)

is largest when

\angle KOL = 90^\circ

Solution:

The nearest point

B

lies on segment

OA

with

OB = \sqrt{13}

and

AB = 4.

Triangles

OKL

and

BKL

share the base

KL,

and their heights are the distances from

O

and

B

to the line through

A.

For any point

P

on line

OA,

that distance is

PA\sin\varphi,

where

\varphi

is the angle between the two lines, so

\frac{[BKL]}{[OKL]} = \frac{AB}{AO} = \frac{4}{4 + \sqrt{13}}.

Since

OK = OL = \sqrt{13},

we have

[OKL] = \frac{13}{2}\sin(\angle KOL) \le \frac{13}{2},

with equality when

\angle KOL = 90^\circ.

Such a chord lies at distance

\sqrt{\frac{13}{2}}

from

O,

which is less than

OA,

so a line through

A

can achieve it.

The maximum area is

\begin{aligned} [BKL] &= \frac{13}{2} \cdot \frac{4}{4 + \sqrt{13}} \\ &= \frac{26}{4 + \sqrt{13}} = \frac{26(4 - \sqrt{13})}{3} \\ &= \frac{104 - 26\sqrt{13}}{3}, \end{aligned}

so

a + b + c + d

= 104 + 26 + 13 + 3

= 146.