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Question: Let A(1,2,1) B(0,0,0) C(2,1,3) and D(1,1,4) be vertices of a tetrahedron ABCD. A plane x+y-z=t cuts ...

Let A(1,2,1) B(0,0,0) C(2,1,3) and D(1,1,4) be vertices of a tetrahedron ABCD. A plane x+y-z=t cuts the tetrahedron into 2 parts whose volumes are in the ratio 1:15. Then, find the sum of squares of possible values of t.

A

0

B

1

C

2

D

4

Answer

2

Explanation

Solution

Here's how to solve the problem:

1. Evaluate the linear form at the vertices:

Given the plane x+y−z=tx + y - z = t, evaluate f(x,y,z)=x+y−zf(x, y, z) = x + y - z at each vertex:

  • f(A)=1+2−1=2f(A) = 1 + 2 - 1 = 2
  • f(B)=0+0−0=0f(B) = 0 + 0 - 0 = 0
  • f(C)=2+1−3=0f(C) = 2 + 1 - 3 = 0
  • f(D)=1+1−4=−2f(D) = 1 + 1 - 4 = -2

2. Barycentric parametrization:

Any point P inside the tetrahedron can be expressed as:

P=B+α(A−B)+β(C−B)+γ(D−B)\mathbf{P} = B + \alpha(A - B) + \beta(C - B) + \gamma(D - B)

where α,β,γ≥0\alpha, \beta, \gamma \ge 0 and α+β+γ≤1\alpha + \beta + \gamma \le 1.

Calculate the differences:

  • A−B=(1,2,1)A - B = (1, 2, 1)
  • C−B=(2,1,3)C - B = (2, 1, 3)
  • D−B=(1,1,4)D - B = (1, 1, 4)

Then,

  • f(A−B)=1+2−1=2f(A - B) = 1 + 2 - 1 = 2
  • f(C−B)=2+1−3=0f(C - B) = 2 + 1 - 3 = 0
  • f(D−B)=1+1−4=−2f(D - B) = 1 + 1 - 4 = -2

So, f(P)=2α−2γ=2(α−γ)f(P) = 2\alpha - 2\gamma = 2(\alpha - \gamma). Let τ=t2\tau = \frac{t}{2}, thus α−γ=τ\alpha - \gamma = \tau.

3. Volume ratios and cumulative volume function:

The cumulative volume function F(t)F(t) (the fraction of the volume where f≤tf \le t) is given by:

  • For t<0t < 0 (τ<0\tau < 0): F(t)=12(3τ+1+3τ2+τ3)F(t) = \frac{1}{2}(3\tau + 1 + 3\tau^2 + \tau^3)
  • For t≥0t \ge 0 (τ≥0\tau \ge 0): F(t)=12(3τ+1−3τ2+τ3)F(t) = \frac{1}{2}(3\tau + 1 - 3\tau^2 + \tau^3)

Since the volumes are in the ratio 1:15, one part is 116\frac{1}{16} of the total volume, and the other is 1516\frac{15}{16}.

  • If t<0t < 0, F(t)=116F(t) = \frac{1}{16}.
  • If t≥0t \ge 0, F(t)=1516F(t) = \frac{15}{16}.

4. Solve for t:

  • Case 1: t<0t < 0

    12(3τ+1+3τ2+τ3)=116\frac{1}{2}(3\tau + 1 + 3\tau^2 + \tau^3) = \frac{1}{16}

    Solving this cubic equation yields t=−1t = -1.

  • Case 2: t≥0t \ge 0

    12(3τ+1−3τ2+τ3)=1516\frac{1}{2}(3\tau + 1 - 3\tau^2 + \tau^3) = \frac{15}{16}

    Solving this cubic equation yields t=1t = 1.

5. Final Answer:

The possible values of tt are -1 and 1. The sum of their squares is (−1)2+(1)2=1+1=2(-1)^2 + (1)^2 = 1 + 1 = 2.

Therefore, the sum of squares of possible values of t is 2.