2020 AMC 12A Problem 24

Question

Suppose that

\triangle ABC

is an equilateral triangle of side length

s,

with the property that there is a unique point

P

inside the triangle such that

AP = 1,

BP = \sqrt{3},

and

CP = 2.

What is

s?

1 + \sqrt{2}

\sqrt{7}

\dfrac{8}{3}

\sqrt{5 + \sqrt{5}}

2\sqrt{2}

For a point at distances

p, q, r

from the vertices of an equilateral triangle of side

s,

the four values satisfy

3(p^4 + q^4 + r^4 + s^4)

= (p^2 + q^2 + r^2 + s^2)^2

Substitute

p^2 = 1,

q^2 = 3,

r^2 = 4

and solve the resulting quadratic in

s^2

A point at distances

p, q, r

from the vertices of an equilateral triangle of side

s

satisfies

3(p^4 + q^4 + r^4 + s^4)

= (p^2 + q^2 + r^2 + s^2)^2.

With

p^2 = 1,

q^2 = 3,

r^2 = 4,

letting

S = s^2

gives

3(26 + S^2) = (8 + S)^2,

so

S^2 - 8S + 7 = 0

and

S = 1

or

S = 7.

A triangle of side

1

cannot contain a point at distance

2

from a vertex, so

S = 7

and

s = \sqrt{7}.

Thus, B is the correct answer.

Problem 24 in other years