2026 AMC 10A Problem 9

Similar triangles have the same shape: the larger one is the smaller one with every side multiplied by the same number, the scale factor. Since the triangles are not congruent, the scale factor is bigger than $1$.

Try an example first. If the small triangle were $9, 12, 15$ and we doubled it, we would get $18, 24, 30$, and the $9$ and the $12$ are gone. Enlarging pushes every side up, so the $9$ and $12$ of the big triangle must come from sides that were smaller before. Let's make that precise.

1. Set up. The smaller triangle has sides $9$, $12$ and a third side; call it $x$. Let $k$ be the scale factor, with $k > 1$. The larger triangle has sides
$$9k, \qquad 12k, \qquad xk.$$

2. Where does the big triangle's $9$ come from? It is not $9k$, because $9k$ is more than $9$. It is not $12k$, because $12k$ is more than $12$, which is already more than $9$. So it must be the third one:
$$xk = 9.$$
(Since $k > 1$, this means $x$ is less than $9$: the third side is the smallest side of the small triangle.)

3. Where does the big triangle's $12$ come from? It is not $12k$, which is more than $12$. It is not $xk$, which we just found is $9$. So it must be
$$9k = 12, \qquad\text{so}\qquad k = \frac{12}{9} = \frac43.$$

4. Find all the sides. The small triangle's third side is $x = 9 \div \tfrac43 = \tfrac{27}{4} = 6.75$. So
- smaller triangle: $6.75,\ 9,\ 12$,
- larger triangle: $9,\ 12,\ 16$ (the last one is $12 \times \tfrac43 = 16$).

5. Check that these really are triangles of the right kind.
- Triangle inequality: $6.75 + 9 = 15.75 > 12$ and $9 + 12 = 21 > 16$, so both triangles exist.
- Each has three different side lengths, so both are scalene.
- Ratios: $9 \div 6.75$, $12 \div 9$ and $16 \div 12$ all equal $\tfrac43$, so they are similar, and they are different sizes.

The longest side of the larger triangle is $16$.

The answer is $\boxed{\textbf{(D)}\ 16}$.

Line up the four numbers $6.75, 9, 12, 16$: each one is $\tfrac43$ times the one before. The smaller triangle uses the first three and the larger triangle uses the last three, so the $9$ and $12$ are shared but sit in different positions. Whenever two similar figures share some lengths, ask which side of one matches which side of the other; a scale factor bigger than $1$ only lets small sides match bigger ones, and that usually leaves just one possibility.

Test the choices by shrinking. Say the larger triangle is $9, 12, L$, where $L$ is its longest side. Shrinking it by the scale factor must give a triangle that still has a $9$ and a $12$.

Shrinking makes every side shorter, so the small triangle's $12$ must come from $L$ (the only side longer than $12$), and its $9$ must come from the $12$.

• For (C), $L = 15$: shrinking $15$ to $12$ is a factor of $\tfrac{15}{12} = \tfrac54$. Then $12$ shrinks to $9.6$ and $9$ shrinks to $7.2$. The small triangle is $7.2, 9.6, 12$, with no side of $9$. So (C) fails.
• For (D), $L = 16$: shrinking $16$ to $12$ is a factor of $\tfrac43$, which takes $12$ to $9$ and $9$ to $6.75$. The small triangle is $6.75, 9, 12$, which has both. So (D) works.