2025 AMC 10B Problem 19

P16-20~6 minPrint

Solid Geometry Ratios, Percents & Averages Add construction lines/points (drop altitudes, extend segments, connect centers)Cut the figure into known shapes (triangles, rectangles, sectors)

A container has a $1\times 1$ square bottom, a $3\times 3$ open square top, and four congruent trapezoidal sides, as shown. Starting when the container is empty, a hose that runs water at a constant rate takes $35$ minutes to fill the container up to the midline of the trapezoids.

How many more minutes will it take to fill the remainder of the container?

Show solution without answering

Key insight

Extend the slanted sides to a point: the container is a pyramid with its tip cut off, and volume below a square of side s grows like s³.

Solution

See the pyramid. The bottom is $1 \times 1$ and the top is $3 \times 3$ , and the four slanted sides are the same. Slice the container level with the ground at any height: you get a square, and its side grows steadily from $1$ at the bottom to $3$ at the top. If you kept the slanted sides going downward, the square would shrink to a point. So the container is a big square pyramid (tip pointing down) with a small pyramid cut off the bottom.

How volume grows in a pyramid. Pyramids of the same shape scale in every direction. If one pyramid's square top has side twice as long as another's, it is also twice as tall, so its volume is $2 \times 2 \times 2 = 8$ times as big. In general, the volume from the tip up to a square of side $s$ is some fixed number times $s^3$ . Call that fixed number $c$ , so the volume is $c \cdot s^3$ .

3. The three levels that matter.
- Bottom of the container: side $1$ , volume below it $c \cdot 1 = c$ .
- The midline, halfway up: the side is halfway between $1$ and $3$ , which is $2$ . Volume below it: $c \cdot 8 = 8c$ .
- Top: side $3$ , volume below it $c \cdot 27 = 27c$ .

4. Water in each half of the container.
- Lower half: $8c - c = 7c$ .
- Upper half: $27c - 8c = 19c$ .

5. Convert to time. The hose runs at a steady rate. The lower half, $7c$ , took $35$ minutes, so each $c$ takes $5$ minutes. The upper half needs
$19 \times 5 = 95 \text{ minutes}.$

The answer is $\boxed{\textbf{(D)}\ 95}$ .

Why this works

A container with a flat bottom that widens evenly is a pyramid (or cone) with its tip cut off. Putting the tip back lets you use one rule: volume grows like the cube of the size. You never need the height or the constant $c$ , only the side lengths $1$ , $2$ , $3$ and their cubes. The same idea answers "how full is a cone-shaped cup filled to half its depth?"

Alternative approach

Use the formula for a pyramid with its tip cut off: volume $= \frac{h}{3}\left(A_1 + \sqrt{A_1 A_2} + A_2\right)$ , where $A_1$ and $A_2$ are the two square areas and $h$ is the height. Each half has the same height $h$ .
- Lower half, areas $1$ and $4$ : $\frac{h}{3}(1 + 2 + 4) = \frac{7h}{3}$ .
- Upper half, areas $4$ and $9$ : $\frac{h}{3}(4 + 6 + 9) = \frac{19h}{3}$ .

The ratio is $7 : 19$ again, so the time is $35 \times \frac{19}{7} = 95$ minutes.

Concise solution

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

The four sides are congruent trapezoids joining a $1 \times 1$ square to a concentric $3 \times 3$ square, so the horizontal cross-section at height $t$ is a square whose side grows linearly from $1$ at the bottom to $3$ at the top. Extending the four faces downwards, the side length shrinks to $0$ at a single point: the container is the part of a square pyramid lying between the cross-sections of side $1$ and side $3$ .

For that pyramid, the portion below the cross-section of side $s$ is similar to the whole pyramid with ratio $s$ , so its volume is
$V(s) = c\,s^3$
for some constant $c$ .

Three levels matter. The bottom is $s = 1$ . The midline of a trapezoid is the segment joining the midpoints of its legs, so filling "up to the midline" means filling to half the height of the container, where the side is the average $\tfrac{1+3}{2} = 2$ . The top is $s = 3$ . Hence

water poured so far: $V(2) - V(1) = c(8 - 1) = 7c$ ,
water still to come: $V(3) - V(2) = c(27 - 8) = 19c$ .

The hose runs at a constant rate, so time is proportional to volume: $7c$ takes $35$ minutes, i.e. $5$ minutes per unit of $c$ , and the remaining $19c$ takes
$19 \cdot 5 = 95 \text{ minutes}.$

The answer is $\boxed{\textbf{(D)}\ 95}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 14: Three-Dimensional Geometry
Introduction to Algebra, Chapter 6: Ratios and Percents
Introduction to Geometry, Chapter 19: Problem Solving Strategies in Geometry (§19.1)
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2, §4.3)