Question

Let $\triangle ABC$ be an equilateral triangle with center O and side length $4\sqrt{3}$. Let P be the segment $\overline{AC}$ such that $CP = \sqrt{6}$. Let $\overline{OP}$ intersect $\overline{AB}$ at Q. Find AQ. Express your form of $\frac{a\sqrt{b}+c\sqrt{d}}{e}$, where a, b, c, d, and e are positive integers such that $b \neq d$ and gcd(a, e) =

Answer 1

$\frac{52\sqrt{3} + 8\sqrt{6}}{23}$

Answer 2

We start by placing the triangle in the coordinate plane.

1. Assign coordinates:

Let

A = (0, 0),

B = (4√3, 0)                    [since AB = 4√3],

C = (2√3, 6).                    [altitude = (4√3)(√3/2) = 6].

2. Find the center O:

O is the centroid:

O = ((0 + 4√3 + 2√3)/3, (0 + 0 + 6)/3) = (2√3, 2).

3. Locate point P on AC with CP = √6:

Parameterize AC: any point is A + t(C – A):

P = (2√3·t, 6t).

Since AC = 4√3, then CP = (1 – t)(4√3) = √6.

Solve for t:

(1 – t) = √6/(4√3) = (√6/√3)/4 = (√2)/4, so t = 1 – (√2)/4.

Thus,

P = (2√3(1 – √2/4), 6(1 – √2/4))

= (2√3 – (√6)/2, 6 – (3√2)/2).

4. Find Q, the intersection of OP with AB:

Line OP passes through O = (2√3,2) and P = (2√3 – (√6)/2, 6 – (3√2)/2). Its direction vector is:

P – O = ( – (√6)/2, [6 – (3√2)/2] – 2).

Calculate the y–component:

6 – (3√2)/2 – 2 = 4 – (3√2)/2 = (8 – 3√2)/2.

Thus, parametric equations for OP are:

x = 2√3 – λ·(√6)/2,

y = 2 + λ·(8 – 3√2)/2.

Since AB is along the x–axis (y = 0), set y = 0:

2 + λ·(8 – 3√2)/2 = 0 → Multiply by 2: 4 + λ(8 – 3√2) = 0,

so λ = –4/(8 – 3√2).

Now, the x–coordinate of Q is:

x = 2√3 – [λ (√6)/2] = 2√3 + (4√6)/(2(8 – 3√2)) = 2√3 + (2√6)/(8 – 3√2).

5. Simplify the expression for AQ:

Rationalize (2√6)/(8 – 3√2) by multiplying numerator and denominator by (8 + 3√2):

(2√6(8 + 3√2))/(64 – 18) = (2√6(8 + 3√2))/(46) = (√6(8 + 3√2))/23.

Also write 2√3 as (46√3)/23.

Thus,

AQ = (46√3 + √6(8 + 3√2))/23.

Expand √6(8 + 3√2):

= 8√6 + 3√12 = 8√6 + 6√3.

So,

AQ = (46√3 + 6√3 + 8√6)/23 = (52√3 + 8√6)/23.