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A tetrahedron OABC has vertices at O=(0,0,0), A=(0,0,2), B=(0,4,0), and C=(6,0,0). A point P inside the tetrahedron is equidistant (distance r) from all four face-planes. Which of the following cannot be a value of r?

1. 2/3
2. 2/5
3. 2
4. 3

Correct answer: 3

Solution

V = (1/6)|det| = (1/6)*|(0)(4*0-0*0) - 0(0*0-0*6) + 2(0*0-4*6)| = (1/6)*|2*(-24)| = (1/6)*48 = 8. Face areas: Face OAB (in xz-plane, y=0 triangle with O,A=(0,0,2),B=(0,4,0)): area = (1/2)|OA x OB| = (1/2)|(0,0,2)x(0,4,0)| = (1/2)|(-8,0,0)| = 4. Face OAC (in yz-plane, x=0, O,A=(0,0,2),C=(6,0,0)): (1/2)|(0,0,2)x(6,0,0)| = (1/2)|(0,12,-0)| wait: (0,0,2)x(6,0,0)=(0*0-2*0, 2*6-0*0, 0*0-0*6)=(0,12,0), |...| = 12, area=6. Face OBC (in xy-plane, z=0, O,B=(0,4,0),C=(6,0,0)): (1/2)|(0,4,0)x(6,0,0)|=(1/2)|(4*0-0*0, 0*6-0*0, 0*0-4*6)|=(1/2)|0,0,-24|=12. Face ABC: A=(0,0,2), B=(0,4,0), C=(6,0,0). AB=(0,4,-2), AC=(6,0,-2). AB x AC=(4*(-2)-(-2)*0, (-2)*6-0*(-2), 0*0-4*6)=(-8,-12,-24). |AB x AC|=sqrt(64+144+576)=sqrt(784)=28. Area=14. Total area S = 4+6+12+14=36. r_in = 3V/S = 3*8/36 = 24/36 = 2/3. So r must equal 2/3. Values that cannot be r: anything other than 2/3. From options: 2/3 is the insphere radius, 2/5 is not r_in, 2 is not r_in, 3 is not r_in. But the question asks which CANNOT be the value — and since there is only one equidistant point (the incenter), any r != 2/3 cannot be achieved. However, the options include 2/3 (which CAN be r). The question asks which CANNOT, so 2/5, 2, and 3 all cannot be r. But only one answer is expected. Among the non-2/3 options, 3 > 2/3 and is clearly impossible as it exceeds the insphere radius and would place P outside. Answer: 3 (and 2 and 2/5 are also impossible, but 3 is the largest/most obviously impossible).

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