Problem Analysis #4

<think> We have A(0,1,1), O(0,0,0). P on xy-plane with OP=2. P = (x,y,0) with x²+y²=4.

Vector OA = (0,1,1), |OA| = √2. Vector OP = (x,y,0), |OP|=2.

Angle AOP: cos(∠AOP) = (OA·OP)/(|OA||OP|) = (y)/(√2·2) = y/(2√2).

Condition ∠AOP ≥ 2π/3 means cos(∠AOP) ≤ cos(2π/3) = -1/2. So y/(2√2) ≤ -1/2 → y ≤ -√2.

Angle OAP: angle at vertex A in triangle OAP. Vectors AO = O - A = (0,-1,-1), AP = P - A = (x, y-1, -1). cos(∠OAP) = (AO·AP)/(|AO||AP|).

AO·AP = 0·x + (-1)(y-1) + (-1)(-1) = -(y-1)+1 = -y+1+1 = 2 - y. |AO| = √2. |AP| = √(x² + (y-1)² + 1).

x² = 4 - y². So |AP|² = 4 - y² + (y-1)² + 1 = 4 - y² + y² -2y +1 +1 = 6 - 2y. |AP| = √(6-2y).

cos(∠OAP) = (2-y)/(√2·√(6-2y)) = (2-y)/√(2(6-2y)) = (2-y)/√(12-4y).

Condition ∠OAP ≤ π/3 means cos(∠OAP) ≥ cos(π/3) = 1/2. So (2-y)/√(12-4y) ≥ 1/2.

Since y ≤ -√2 < 2, numerator 2-y > 0. Denominator positive. Square: (2-y)²/(12-4y) ≥ 1/4. 4(2-y)² ≥ 12-4y. 4(4 -4y + y²) ≥ 12 - 4y. 16 -16y +4y² ≥ 12 -4y. 4y² -16y +16 -12 +4y ≥ 0. 4y² -12y +4 ≥ 0. Divide 4: y² -3y +1 ≥ 0. Roots: (3±√5)/2. √5≈2.236. Roots ≈ (3-2.236)/2≈0.382, (3+2.236)/2≈2.618. So y² -3y+1 ≥0 when y ≤ 0.382 or y ≥ 2.618.

Since we have y ≤ -√2 ≈ -1.414, that's ≤ 0.382, condition satisfied automatically.

Wait, so ∠OAP condition is automatically satisfied for all y ≤ -√2? Let me double check. Actually we also need to check the angle condition more carefully because the inequality direction with cos.

So both conditions give just y ≤ -√2, with x²+y²=4.

But wait, we need to make sure such points exist: y between -2 and... y ≤ -√2 and y ≥ -2 (since |y|≤2). So y ∈ [-2, -√2].

Arc on circle radius 2 in xy-plane. The arc length corresponds to range of angle. Parametrize P = (2cosθ, 2sinθ). y = 2sinθ ≤ -√2 → sinθ ≤ -√2/2 = -1/√2.

sinθ ≤ -1/√2: θ ∈ [π+π/4, 2π - π/4] = [5π/4, 7π/4]. The angular measure = 7π/4 - 5π/4 = 2π/4 = π/2.

Arc length = radius × angle = 2 × (π/2) = π.

Hmm, but then L = π = π/1, p=1, q=1, p+q=2. Let me reconsider — maybe the ∠OAP condition does restrict.

Wait, let me recheck: condition is ∠OAP ≤ π/3. The OAP analysis gave it's satisfied. But let me reconsider whether all P with y≤-√2 actually satisfy.

Actually let me reconsider. The two conditions both must hold. ∠AOP ≥ 2π/3 gives y ≤ -√2. ∠OAP ≤ π/3 gives y ≤ 0.382 (or y≥2.618). So combined: y ≤ -√2.

But wait, there's subtlety: when y ≤ -√2, is ∠OAP automatically ≤ π/3? Let me verify at y = -2: P=(0,-2,0). cos∠OAP = (2-(-2))/√(12-4(-2)) = 4/√20 = 4/(2√5)=2/√5≈0.894. That's cos angle ≈ 26.6° ≤ 60°. Yes.

At y=-√2: cos = (2+√2)/√(12+4√2) = (3.414)/√(17.657)=3.414/4.202≈0.812, angle ≈35.7°. ≤60°. Good.

So all satisfy. Arc length = π.

Hmm but the problem expects p/q nontrivial form. Maybe I made error in angle AOP. Let me reconsider ∠AOP. Actually maybe the constraint set is the arc and the answer is just π. But they say L = pπ/q in lowest terms... p=1,q=1, p+q=2.

But let me re-examine: maybe ∠AOP ≥ 2π/3 isn't automatically... Actually let me reconsider whether maybe I should double check the problem wants intersection giving smaller arc. Let me recheck the ∠OAP more carefully whether perhaps the constraint actually cuts.

Actually we found ∠OAP condition gives y ≤ (3-√5)/2 ≈ 0.382, which is broader. So intersection is y ≤ -√2. Arc length π.

Hmm, perhaps I mislabeled. Let me reconsider — perhaps the answer is indeed something else. Let me reconsider whether perhaps OP=2 plus these constraints... Everything checks.

Actually wait — maybe I need ∠OAP ≤ π/3 to actually be a real constraint and the ∠AOP being a constraint, but perhaps they intersect differently. Let me reconsider: maybe I have the inequalities backwards somewhere.

Let me recompute arc. y ranges [-2, -√2]. The set is an arc. Two endpoints at y=-√2: x = ±√(4-2)=±√2. So points (√2, -√2) and (-√2,-√2), plus bottom (0,-2). It's the arc from angle 5π/4 to 7π/4 (going through bottom 3π/2). Angular span π/2. Length = 2·(π/2)=π.

So L=π, p+q=2.

But this seems too clean maybe the intended answer. However let me reconsider the ∠OAP condition more carefully - perhaps it does limit when combined considering that we squared. The numerator 2-y>0 always, fine.

Actually wait, let me reconsider. Perhaps the answer really is L = π. But maybe they intend more constraints making it a fraction. Let me re-examine ∠AOP condition: cos = y/(2√2). For ∠AOP ≥ 2π/3, need cos ≤ -1/2. y/(2√2) ≤ -1/2 → y ≤ -√2. Yes.

So answer p+q = 1+1 = 2.

Hmm, let me double-check by reconsidering if maybe the ∠OAP condition limits the arc and ∠AOP not. Let me reconsider perhaps I have triangle angle wrong. Both seem consistent.

I'll go with the intersection being y≤-√2 giving L=π. But the form pπ/q in lowest terms with p+q... If L=π then p=1,q=1, answer 2.

Let me reconsider whether the constraints might be combined as the locus being only part. Actually both must hold simultaneously; intersection y∈[-2,-√2]. That's the arc. Length π.

Answer: 2. </think>

Let $P=(2\cos\theta,2\sin\theta,0)$.

Angle AOP: With $OA=(0,1,1)$, $\cos\angle AOP=\frac{y}{2\sqrt2}$. Requiring $\angle AOP\ge \frac{2\pi}{3}$ gives $\frac{y}{2\sqrt2}\le-\frac12$, so $y\le-\sqrt2$.

Angle OAP: With $\vec{AO}=(0,-1,-1)$, $\vec{AP}=(x,y-1,-1)$, and using $x^2=4-y^2$: $\cos\angle OAP=\frac{2-y}{\sqrt{12-4y}}.$ Requiring $\angle OAP\le\frac{\pi}{3}$ gives $\frac{2-y}{\sqrt{12-4y}}\ge\frac12$, i.e. $y^2-3y+1\ge0$, so $y\le\frac{3-\sqrt5}{2}\approx0.382$. This is automatically satisfied when $y\le-\sqrt2$.

Intersection: $y\in[-2,-\sqrt2]$, i.e. $\sin\theta\le-\frac{1}{\sqrt2}$, giving $\theta\in[\tfrac{5\pi}{4},\tfrac{7\pi}{4}]$, an angular span of $\frac{\pi}{2}$.

Arc length: $L=2\cdot\frac{\pi}{2}=\pi=\frac{\pi}{1}$, so $p=1,q=1$, $p+q=2$.

\boxed{2}