2017 AMC 10A Problem 10

P1-10~3 minPrint

Triangles: Area & Pythagorean Basic Counting Bound the quantity above/below or estimate to pin it down

Joy has $30$ thin rods, one each of every integer length from $1$ cm through $30$ cm. She places the rods with lengths $3$ cm, $7$ cm, and $15$ cm on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?

Show solution without answering

Key insight

The longest side must be shorter than the other three combined. That allows lengths 6 through 24, minus the rods 7 and 15 already used.

Solution

For three sticks to make a triangle, the longest must be shorter than the other two put together. Four sticks work the same way: the longest must be shorter than the other three put together. Otherwise the other three cannot reach around to close the shape.

Call the length of the fourth rod $d$ .

$d$ cannot be too long. If $d$ is the longest side, it must be shorter than $3 + 7 + 15 = 25$ . So $d$ is at most $24$ .
$d$ cannot be too short. If $15$ is the longest side, it must be shorter than $3 + 7 + d = 10 + d$ . So $d$ must be more than $5$ , that is, at least $6$ .
So $d$ can be any whole number from $6$ to $24$ . That is $24 - 6 + 1 = 19$ lengths.
Joy has only one rod of each length, and the $7$ cm and $15$ cm rods are already on the table. Both are in the list from $6$ to $24$ , so remove them: $19 - 2 = 17$ .

The answer is $\boxed{\textbf{(B)}\ 17}$ .

Why this works

The "longest side is less than the rest" rule traps the missing side from both ends: it cannot be longer than the others combined, and it cannot be so short that $15$ becomes too long. After you find the range, read the problem once more for numbers that are not allowed. "One each of every length" is the clue that $7$ and $15$ are gone.

Concise solution

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

Four lengths form a quadrilateral with positive area exactly when the longest one is strictly less than the sum of the other three (the same idea as the triangle inequality, one side more).

Let the fourth rod have length $d$ .

If $d$ is the longest side, we need $d < 3 + 7 + 15 = 25$ , so $d \le 24$ .

Otherwise $15$ is the longest side, and we need $15 < 3 + 7 + d$ , so $d > 5$ , i.e. $d \ge 6$ .

So $d$ can be any integer from $6$ to $24$ : that is $24 - 6 + 1 = 19$ values. But the rods of length $7$ and $15$ are already in use, so they cannot be chosen again. That leaves $19 - 2 = 17$ .

The answer is $\boxed{\textbf{(B)}\ 17}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Geometry, Chapter 6: Right Triangles (§6.1, §6.2, §6.3)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques