AMC 10 · 2008 · #18

Grade 10 geometry-3d

Archetype Solid Volume by Decomposition └ Sub-archetype Fraction or Difference of a Bounding Solid

pythagorean-theoremsystems-of-equationsvolume-pyramid convert-to-algebraspatial-visualization ↑ Prerequisites: pythagorean-theorem

📏 Long solution 💡 2 insights

Problem

A triangle with sides 5, 6 and 7 has each vertex on a different positive coordinate axis. Find the volume of the tetrahedron it makes with the origin.

Pick an answer.

(A)

$\sqrt{85}$

(B)

$\sqrt{90}$

(C)

$\sqrt{95}$

(D)

(E)

$\sqrt{105}$

How to solve

Strategy Introduce a Variable

Nothing in the problem tells you how far out each vertex sits, so name those three distances and let the geometry pin them down. Because the axes meet at right angles, every side of the triangle is the hypotenuse of a right triangle whose legs are two of the named distances, which turns the picture into three Pythagorean equations. Those equations look hard in the distances themselves but are plain linear equations in their squares, so re-reading the system with the squares as the unknowns is the move that cracks it. Finally, the same right angles make the volume formula easy, since one edge doubles as the height.

1STEP 1

Put the triangle on the axes

Each side becomes a right triangle at the origin.

A=(a,0,0), B=(0,b,0), C=(0,0,c), a,b,c ＞ 0

2STEP 2

Write three Pythagoras equations

Treating the squares as unknowns makes the system linear.

a²+b²=5²=25, b²+c²=6²=36, c²+a²=7²=49

3STEP 3

Make the squares the unknowns

Adding all three gives their total at once.

(a²+b²)+(b²+c²)+(c²+a²)=2(a²+b²+c²)=110 → a²+b²+c²=55

4STEP 4

Peel off each square

One subtraction each peels off every square.

c²=55-25=30, a²=55-36=19, b²=55-49=6

5STEP 5

Check the tetrahedron exists

All three come out positive, so the tetrahedron exists.

a²=(25+49-36)/2=19 ＞ 0, 5²+6²=61 ＞ 49=7² (triangle is acute)

6STEP 6

The labeling does not matter

The labelling does not matter, since the set of squares is the same.

{a²,b²,c²}={55-25, 55-36, 55-49}={30,19,6} in every labeling

7STEP 7

Use the right corner as the height

The right corner makes the volume a plain product over six.

V=1/3·(ab/2) · c=abc/6

8STEP 8

Multiply under one square root

Multiplying under one root gives √(95), choice (C).

V=abc/6=√(a²b²c²)/6=(√(19 · 6 · 30))/6=√(3420)/6=6√(95)/6=√(95)

Answer

√(95)

The size is sensible: the tetrahedron sits in the box with edges sqrt(19), sqrt(6), sqrt(30), whose volume is sqrt(3420), about 58.5, and a corner tetrahedron should be exactly one sixth of that box, which is about 9.75, matching sqrt(95). Note that estimating alone cannot settle this problem, since the five choices are about 9.22, 9.49, 9.75, 10, and 10.25, all within one unit of each other, so the exact radical is required. A stronger, independent check ties the three computed edges back to the given triangle without ever touching volume: for a right-angle corner, the square of the area of the slanted face equals the sum of the squares of the areas of the three right-triangle faces at O. Those three areas are ab/2, bc/2, ca/2, and the sum of their squares is (19 · 6+6 · 30+30 · 19)/4=864/4=216. Meanwhile Heron's formula on the 5-6-7 triangle gives area sqrt(9 x 4 x 3 x 2) = 6 sqrt(6), whose square is also 216. The two agree, so a squared = 19, b squared = 6, c squared = 30 are confirmed by a route that used only the triangle's area.

💡Key takeaway

When the unknowns only ever show up squared, treat the squares as your unknowns and the tangled system turns into simple adding and subtracting.

Put the triangle on the axes
Write three Pythagoras equations
Make the squares the unknowns
Peel off each square
Check the tetrahedron exists
The labeling does not matter
Use the right corner as the height
Multiply under one square root