2020 AMC 12B Problem 10

Attempt Problem 10 of the 2020 AMC 12B 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 2020 AMC 12B solutions, or check the answer key.

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10.

In unit square

ABCD,

the inscribed circle

\omega

intersects

\overline{CD}

at

M,

and

\overline{AM}

intersects

\omega

at a point

P

different from

M.

What is

AP?

\dfrac{\sqrt{5}}{12}

\dfrac{\sqrt{5}}{10}

\dfrac{\sqrt{5}}{9}

\dfrac{\sqrt{5}}{8}

\dfrac{2\sqrt{5}}{15}

Answer: B

Concepts:coordinate geometry circle power of a point

Difficulty rating: 1560

Small hint:

Place

A = (0, 0),

so the circle has center

\left(\tfrac12, \tfrac12\right)

and

M = \left(\tfrac12, 1\right)

Big hint:

Substitute the line

AM

into the circle’s equation; the two roots give

M

and

P

Solution:

Let

A = (0, 0),

B = (1, 0),

C = (1, 1),

D = (0, 1).

The inscribed circle has center

\left(\tfrac12, \tfrac12\right)

and radius

\tfrac12,

touching

\overline{CD}

at

M = \left(\tfrac12, 1\right).

Line

AM

is

y = 2x.

Substituting into

\left(x - \tfrac12\right)^2 + \left(y - \tfrac12\right)^2 = \tfrac14

gives

20x^2 - 12x + 1 = 0,

with roots

x = \tfrac12

(point

M

) and

x = \tfrac{1}{10}

(point

P

).

So

P = \left(\tfrac{1}{10}, \tfrac15\right)

and

AP = \sqrt{\left(\tfrac{1}{10}\right)^2 + \left(\tfrac15\right)^2} = \frac{\sqrt5}{10}.

Thus, the correct answer is B.

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