2015 AMC 10A Problem 21

P21-25~7 minPrint

Solid Geometry Triangles: Area & Pythagorean Add construction lines/points (drop altitudes, extend segments, connect centers)Set up the equation/formula and compute; no special trick needed

Tetrahedron $ABCD$ has $AB=5$ , $AC=3$ , $BC=4$ , $BD=4$ , $AD=3$ , and $CD=\tfrac{12}5\sqrt2$ . What is the volume of the tetrahedron?

Show solution without answering

Key insight

Faces ABC and ABD are matching 3-4-5 right triangles on side AB. Their heights to AB meet at one point, and CD shows they are perpendicular.

Solution

For volume we need a base and a height that is truly perpendicular to the base. None of the given edges is that height, so we build it.

1. Spot the right triangles. $3^2 + 4^2 = 9 + 16 = 25 = 5^2$ . So face $ABC$ (sides $3, 4, 5$ ) has a right angle at $C$ , and face $ABD$ (sides $3, 4, 5$ ) has a right angle at $D$ . The two faces are the same triangle, joined along the long side $AB$ .
2. Height of one face. In the $3$ - $4$ - $5$ triangle $ABC$ , drop a perpendicular from $C$ to $AB$ and call its foot $F$ . The area can be worked out two ways: $\frac{1}{2} \cdot 3 \cdot 4 = 6$ , and $\frac{1}{2} \cdot 5 \cdot CF$ . So $CF = \frac{12}{5}$ .
3. Where $F$ lands. In right triangle $ACF$ , $AF^2 = 3^2 - \left(\frac{12}{5}\right)^2 = 9 - \frac{144}{25} = \frac{81}{25}$ , so $AF = \frac{9}{5}$ .
4. The other face. Triangle $ABD$ has $AD = 3$ and $BD = 4$ , exactly like $AC$ and $BC$ . So the same calculation gives a perpendicular from $D$ of length $\frac{12}{5}$ , landing $\frac{9}{5}$ from $A$ : at the same point $F$ .
5. The angle between the two heights. Look at triangle $CFD$ : $CF = DF = \frac{12}{5}$ and $CD = \frac{12}{5}\sqrt{2}$ . Since
$\left(\tfrac{12}{5}\right)^2 + \left(\tfrac{12}{5}\right)^2 = 2\left(\tfrac{12}{5}\right)^2 = \left(\tfrac{12}{5}\sqrt{2}\right)^2,$
the Pythagorean theorem works backwards and the angle at $F$ is $90^\circ$ .
6. So $DF$ is the height. $DF$ is perpendicular to $AB$ (step 4) and to $CF$ (step 5). Both of those lines lie flat in the base $ABC$ and cross each other at $F$ , so $DF$ stands straight up from the base.
7. Volume. A pyramid's volume is $\frac{1}{3} \times \text{base area} \times \text{height}$ :
$V = \frac{1}{3} \times 6 \times \frac{12}{5} = \frac{24}{5}.$

The answer is $\boxed{\textbf{(C)}\ \dfrac{24}{5}}$ .

Why this works

The strange length $\frac{12}{5}\sqrt{2}$ is a clue. $\frac{12}{5}$ is the height of a $3$ - $4$ - $5$ triangle onto its long side, and a factor $\sqrt{2}$ is what the long side of an isosceles right triangle has. So the two faces meet at a right angle, like a book opened to $90^\circ$ . When a problem gives an odd length, ask which right triangle would produce it.

Alternative approach

Cut the solid with the flat triangle $CFD$ . That triangle is perpendicular to $AB$ and has area $\frac{1}{2} \cdot \frac{12}{5} \cdot \frac{12}{5} = \frac{72}{25}$ . It splits the solid into two pyramids, one with tip $A$ and one with tip $B$ , whose heights $AF$ and $FB$ add up to $AB = 5$ . So
$V = \frac{1}{3} \cdot \frac{72}{25} \cdot AF + \frac{1}{3} \cdot \frac{72}{25} \cdot FB = \frac{1}{3} \cdot \frac{72}{25} \cdot 5 = \frac{24}{5}.$

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Since $3^2 + 4^2 = 5^2$ , face $ABC$ is a right triangle with the right angle at $C$ , and face $ABD$ is a right triangle with the right angle at $D$ . The two faces are congruent and share the hypotenuse $AB$ .

Let $F$ be the foot of the perpendicular from $C$ to $AB$ . In a $3$ - $4$ - $5$ triangle the altitude to the hypotenuse is $\frac{3 \cdot 4}{5} = \frac{12}{5}$ , and it lands at distance $\frac{AC^2}{AB} = \frac{9}{5}$ from $A$ . Triangle $ABD$ has the same side lengths in the same positions ( $AD = AC$ , $BD = BC$ ), so the perpendicular from $D$ to $AB$ has the same length $\frac{12}{5}$ and lands at the same point $F$ .

Now $CF = DF = \frac{12}{5}$ and $CD = \frac{12}{5}\sqrt2$ , so $CF^2 + DF^2 = CD^2$ and $\angle CFD = 90^\circ$ . The segment $DF$ is therefore perpendicular to both $AB$ and $CF$ , two intersecting lines of the plane $ABC$ , so $DF$ is perpendicular to that plane: it is the height of the tetrahedron over base $ABC$ .

The base has area $\frac12 \cdot 3 \cdot 4 = 6$ , so
$V = \frac13 \cdot 6 \cdot \frac{12}{5} = \frac{24}{5}.$

The answer is $\boxed{\textbf{(C)}\ \dfrac{24}{5}}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 14: Three-Dimensional Geometry
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 19: Problem Solving Strategies in Geometry (§19.1)
Introduction to Geometry, Chapter 15: Curved Surfaces