Problem Analysis #8

<think> Points: A(1,0,0), B(0,1,0), C(0,0,1). D = midpoint of AC = (1/2, 0, 1/2).

Triangle ABD with vertices A(1,0,0), B(0,1,0), D(1/2,0,1/2).

Rotate about z-axis. We need volume of solid of revolution.

For rotation about z-axis, a point (x,y,z) maps to a circle of radius r = sqrt(x²+y²) at height z. The solid of revolution is the union of all circles traced by points in the triangle.

For each height z, the set of points in the triangle at that height z forms a line segment (cross-section). Rotating gives an annulus (or disk) with radii from r_min(z) to r_max(z).

We need to find for each z, the range of distances r = sqrt(x²+y²) of points in the triangle at height z.

Let me parametrize the triangle. Vertices: A=(1,0,0), B=(0,1,0), D=(1/2,0,1/2).

z ranges from 0 to 1/2 (since A,B at z=0, D at z=1/2).

At height z, the cross section of triangle. The triangle plane: find it.

Edges:

• AB: from (1,0,0) to (0,1,0), z=0.
• AD: from (1,0,0) to (1/2,0,1/2). On this edge y=0.
• BD: from (0,1,0) to (1/2,0,1/2).

At a given height z in [0,1/2], the cross-section is a segment connecting points on edges AD and BD (both go from z=0 to z=1/2).

On AD: parametrize t∈[0,1]: A + t(D-A) = (1 - t/2, 0, t/2). z = t/2, so t = 2z. Point: (1 - z, 0, z).

On BD: B + s(D-B) = (s/2, 1-s, s/2). z = s/2, s=2z. Point: (z, 1-2z, z).

So at height z, the cross-section is segment from P1=(1-z, 0, z) to P2=(z, 1-2z, z).

We rotate this segment about z-axis. The distance from z-axis: r = sqrt(x²+y²).

Points on segment: parametrize u∈[0,1]: x = (1-z) + u(z - (1-z)) = (1-z) + u(2z-1) y = 0 + u(1-2z) = u(1-2z) r²(u) = x² + y².

We need min and max of r over u∈[0,1].

The rotated cross-section at height z is annulus with inner radius r_min and outer r_max (assuming segment doesn't pass through... actually if r_min could be 0).

Volume = ∫ π(r_max² - r_min²) dz over z from 0 to 1/2.

Let me compute r²(u).

Let a = 1-z, b = 2z-1, c = 1-2z = -b. x = a + u·b y = u·c = -u·b... wait c=1-2z, b=2z-1=-c. So b=-c.

x = a + u·b = (1-z) + u(2z-1) y = u(1-2z) = u·c where c=1-2z.

r² = [(1-z)+u(2z-1)]² + [u(1-2z)]²

Let me denote. Let’s set: term x = (1-z) - u(1-2z) since 2z-1 = -(1-2z). So x = (1-z) - u(1-2z). Let k = 1-2z (≥0 for z≤1/2), and m = 1-z. x = m - u·k y = u·k

r² = (m - uk)² + (uk)² = m² - 2muk + u²k² + u²k² = m² - 2muk + 2u²k².

This is quadratic in u: f(u) = 2k² u² - 2mk u + m².

Min at u* = (2mk)/(2·2k²) = m/(2k).

f(u*) = m² - 2mk·(m/(2k)) + 2k²·(m/(2k))² = m² - m² + 2k²·m²/(4k²) = m²/2.

So minimum value (over all u) is m²/2, achieved at u* = m/(2k).

But u must be in [0,1]. Check if u* ∈[0,1].

u* = m/(2k) = (1-z)/(2(1-2z)).

At z=0: u*=1/2 ∈[0,1]. Good. At z→1/2: k→0, u*→∞. So for some z, u* exceeds 1.

Find when u*=1: (1-z)/(2(1-2z))=1 → 1-z = 2(1-2z)=2-4z → 1-z = 2-4z → 3z =1 → z=1/3.

For z<1/3, u*<1 so interior minimum exists (r_min² = m²/2). For z≥1/3, u*≥1, so f is decreasing on [0,1]? f is parabola opening up, min at u*>1, so on [0,1] f decreasing, min at u=1.

Let me evaluate endpoints: f(0) = m² = (1-z)². This is r² at P1 = (1-z,0,z): indeed (1-z)². Correct. f(1) = 2k² - 2mk + m². At P2=(z,1-2z,z): r² = z²+(1-2z)². Let's check: 2k²-2mk+m² with k=1-2z, m=1-z. 2(1-2z)² - 2(1-z)(1-2z) + (1-z)². Hmm let me just trust f(1) = m² -2mk+2k².

Actually let's verify P2: x=z? P2 x-coord = z. r² = z² + (1-2z)². Let me compute f(1): x at u=1: m-k = (1-z)-(1-2z)=z. Good x=z. y=k=1-2z. So r²= z²+(1-2z)². Yes matches.

So r_max² and r_min²:

For each z, over u∈[0,1], the range of r². The maximum is the larger of f(0) and f(1). The minimum is either f(u*) (if u*∈[0,1]) or the smaller endpoint.

The rotated solid at height z: union of circles of radius r(u) for u∈[0,1]. Since r(u) takes all values between min and max continuously (it's continuous), the annulus has inner radius = min r, outer radius = max r.

Volume contribution = π(r_max² - r_min²) dz.

Case 1: z ∈ [0, 1/3]. u*∈[0,1], r_min² = m²/2 = (1-z)²/2. r_max² = max(f(0), f(1)). f(0)=m²=(1-z)². f(1)= z²+(1-2z)². Compare (1-z)² vs z²+(1-2z)². (1-z)² = 1-2z+z². z²+(1-2z)² = z² + 1-4z+4z² = 5z²-4z+1. Difference f(0)-f(1) = (1-2z+z²) - (5z²-4z+1) = -4z² +2z = 2z(1-2z) ≥0 for z∈[0,1/2]. So f(0) ≥ f(1). Thus r_max² = f(0) = (1-z)².

So for z∈[0,1/3]: π(r_max² - r_min²) = π[(1-z)² - (1-z)²/2] = π(1-z)²/2.

Case 2: z∈[1/3, 1/2]. u*≥1, f decreasing on [0,1]. So min at u=1: r_min²=f(1)=z²+(1-2z)²; max at u=0: r_max²=f(0)=(1-z)².

Contribution = π[(1-z)² - (z²+(1-2z)²)] = π·[2z(1-2z)] = π·(2z-4z²).

Wait that's f(0)-f(1)=2z(1-2z). Yes.

Now integrate.

V = ∫₀^{1/3} π(1-z)²/2 dz + ∫_{1/3}^{1/2} π(2z-4z²) dz.

Compute I1 = (π/2)∫₀^{1/3} (1-z)² dz. ∫(1-z)² dz = -(1-z)³/3. Eval 0 to 1/3: [-(1-1/3)³/3] - [-(1)³/3] = -( (2/3)³)/3 + 1/3 = -(8/27)/3 + 1/3 = -8/81 + 27/81 = 19/81. I1 = (π/2)(19/81) = 19π/162.

I2 = π∫_{1/3}^{1/2} (2z-4z²) dz. ∫(2z-4z²)dz = z² - (4/3)z³. At 1/2: (1/4) - (4/3)(1/8) = 1/4 - 1/6 = 3/12-2/12=1/12. At 1/3: (1/9) - (4/3)(1/27) = 1/9 - 4/81 = 9/81-4/81=5/81. Difference: 1/12 - 5/81. LCD 324: 1/12=27/324, 5/81=20/324. Diff=7/324. I2 = 7π/324.

V = 19π/162 + 7π/324 = 38π/324 + 7π/324 = 45π/324. Simplify: 45/324 = 5/36. V = 5π/36.

So p=5, q=36, p+q=41.

Let me double check whether the solid might have a hole or filled center. At z∈[0,1/3], inner radius r_min = (1-z)/√2 >0, so there's a hole. Hmm but is that right? The annulus formed by rotating a segment that doesn't reach the axis. The minimum distance of the segment to the z-axis is (1-z)/√2. So yes hole.

But wait—we should consider the entire solid as union over z. The cross-sections at different z are separate. Each z gives an annulus. Volume = ∫π(r_max²-r_min²)dz. That's correct as long as the solid at each height is exactly that annulus (it is, since rotating produces full annulus).

So V=5π/36, answer 41. </think>

Setting up the rotation.

With $A(1,0,0)$, $B(0,1,0)$, $D=(\tfrac12,0,\tfrac12)$, the triangle's cross-section at height $z\in[0,\tfrac12]$ is the segment between $P_1=(1-z,\,0,\,z)\ \text{(on }AD),\qquad P_2=(z,\,1-2z,\,z)\ \text{(on }BD).$

Parametrize with $u\in[0,1]$, letting $k=1-2z,\ m=1-z$: $x=m-uk,\quad y=uk,\qquad r^2(u)=m^2-2mku+2k^2u^2.$

Min/max of $r^2$. This parabola in $u$ has minimum at $u^*=\dfrac{m}{2k}$ with value $\dfrac{m^2}{2}$.

• $u^*\le 1 \iff z\le\tfrac13$.
• Endpoints: $f(0)=(1-z)^2$, $f(1)=z^2+(1-2z)^2$, and $f(0)-f(1)=2z(1-2z)\ge0$, so $f(0)$ is the larger.

Case $z\in[0,\tfrac13]$: $r_{\min}^2=\tfrac{(1-z)^2}{2}$, $r_{\max}^2=(1-z)^2$. $\Delta = (1-z)^2-\tfrac{(1-z)^2}{2}=\tfrac{(1-z)^2}{2}.$

Case $z\in[\tfrac13,\tfrac12]$: $f$ decreasing, so $r_{\min}^2=f(1)$, $r_{\max}^2=f(0)$. $\Delta = 2z(1-2z)=2z-4z^2.$

Volume. $V=\frac{\pi}{2}\int_0^{1/3}(1-z)^2\,dz+\pi\int_{1/3}^{1/2}(2z-4z^2)\,dz.$

First integral: $\int_0^{1/3}(1-z)^2dz=\left[-\tfrac{(1-z)^3}{3}\right]_0^{1/3}=\frac{27-8}{81}=\frac{19}{81},$ so $I_1=\dfrac{19\pi}{162}$.

Second integral: $\Big[z^2-\tfrac43z^3\Big]_{1/3}^{1/2}=\frac{1}{12}-\frac{5}{81}=\frac{7}{324},$ so $I_2=\dfrac{7\pi}{324}$.

$V=\frac{19\pi}{162}+\frac{7\pi}{324}=\frac{38\pi+7\pi}{324}=\frac{45\pi}{324}=\frac{5\pi}{36}.$

Thus $p=5,\ q=36$, giving $p+q=41$.

\boxed{41}