Trigonometry & Inverse Trigonometry Question - None

Question

The inradius of $\triangle ABC$ is $100\sqrt{3}$ and the circumradius is $200\sqrt{3}$. Consider the line perpendicular to plane $ABC$ through the circumcenter of $\triangle ABC$. Note that $P, Q, O$ must lie on that line to be equidistant from each of the triangle's vertices. Also, note that since $P, Q, O$ are collinear, and $OP = OQ$, we must have $O$ is the midpoint of $PQ$. Now, Let $K$ be the circumcenter of $\triangle ABC$, and $L$ be the foot of the altitude from $A$ to $BC$. We must have $\tan(\angle KLP + \angle QLK) = \tan(120°)$. Setting $KP = x$ and $KQ = y$, assuming WLOG $x > y$, we must have $$\tan(120°) = -\sqrt{3} = \frac{\dfrac{x+y}{100\sqrt{3}}}{\dfrac{30000 - xy}{30000}}.$$ Thus $100(x+y) = xy - 30000$. Also, $\left(\dfrac{x+y}{2}\right)^2 = \left(\dfrac{x-y}{2}\right)^2 + 120000$ by the Pythagorean theorem, so $xy = 120000$, and substituting, $90000 = 100(x+y)$, or $x + y = 900$. The desired answer is $\dfrac{x+y}{2}$.

Accepted Answer

Step 1: Recognize that P, Q, O are collinear on the perpendicular to plane ABC through circumcenter K, with O as midpoint of PQ. This means OP = OQ and KP + KQ = 2·OK.

Step 2: Apply the tangent addition formula: tan(∠KLP + ∠QLK) = tan(120°) = -√3. Using the given formula with inradius r = 100√3 and circumradius R = 200√3, we get: -√3 = [(x+y)/(100√3)] / [(30000 - xy)/30000]

Step 3: Simplify the tangent equation: 100(x+y) = xy - 30000, which gives xy = 100(x+y) + 30000.

Step 4: Apply Pythagorean theorem from the midpoint condition: ((x+y)/2)² = ((x-y)/2)² + 120000. Expanding: (x+y)²/4 - (x-y)²/4 = 120000, which simplifies to xy = 120000.

Step 5: Substitute xy = 120000 into the equation from Step 3: 120000 = 100(x+y) + 30000, so x+y = 900.

Step 6: The desired answer is (x+y)/2 = 900/2 = 450

∴ Answer: 450

Question:

Step-by-Step Solution

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