2026 AMC 10A Problem 14

Three lengths form a triangle with positive area exactly when each length is less than the sum of the other two (the triangle inequality). If one length equals the sum of the other two, the "triangle" is squashed flat and has zero area, so the inequalities must be strict. Write out all three, plus the fact that lengths are positive.

1. The sides must be positive. $c = 4t + 4$ is positive only when $t > -1$, so the integer $t$ is at least $0$. For $t \ge 0$, the other two sides $a = t^2 + 1$ and $b = 5t + 7$ are positive as well.

2. First inequality: $a + b > c$.
$$t^2 + 1 + 5t + 7 > 4t + 4, \quad\text{which simplifies to}\quad t^2 + t + 4 > 0.$$
For $t \ge 0$ the left side is at least $4$, so this is always true. It rules nothing out.

3. Second inequality: $a + c > b$.
$$t^2 + 1 + 4t + 4 > 5t + 7, \quad\text{which simplifies to}\quad t^2 - t - 2 > 0.$$
This factors as $(t - 2)(t + 1) > 0$.
Since $t \ge 0$, the factor $t + 1$ is positive. So the product is positive exactly when $t - 2 > 0$, that is, $t > 2$.

4. Third inequality: $b + c > a$.
$$5t + 7 + 4t + 4 > t^2 + 1, \quad\text{which simplifies to}\quad t^2 - 9t - 10 < 0.$$
This factors as $(t - 10)(t + 1) < 0$.
Again $t + 1$ is positive, so the product is negative exactly when $t - 10 < 0$, that is, $t < 10$.

1. Combine. We need $2 < t < 10$, so $t = 3, 4, 5, 6, 7, 8, 9$. That is $7$ values.

1. Check the edges with numbers. It is enough to compare the longest side with the sum of the other two:

Both edge values $t = 2$ and $t = 10$ give flat triangles, which is exactly why the problem says "positive area."

The answer is $\boxed{\textbf{(B)}\ 7}$.

The triangle inequality turns a geometry condition into three algebra conditions. When the sides are expressions in $t$, each inequality cuts the allowed values of $t$ down, and only the strongest ones matter. Here the square $t^2$ grows much faster than the straight-line expressions, so for large $t$ the side $a$ becomes too long, and for small $t$ the side $b$ is too long. That is why the answer is a finite range. Always test the boundary values in a quick table: they are where "greater than" and "greater than or equal to" give different answers.

Since only integers are allowed, you can simply make a table. Start at $t = 0$ (the smallest value with all sides positive) and compare the longest side with the sum of the other two:

Once $a$ passes $b + c$ it stays ahead, because $a$ grows by more and more each step while $b + c$ grows by only $9$. So the values that work are $t = 3$ through $9$, which is $7$ values.