In any triangle, the length of the longest side is less than half of the perimeter. All triangles with perimeter 57 and integer side lengths $x, y, z$, such that $x<y<z$ are constructed. How many such triangles are there?
(A) 68
(B) 61
(C) 75
(D) 56
(E) 27
The length of the longest side, $z$, is less than half of the perimeter 57 .

Thus, $z<\frac{57}{2}$ or $z<28 \frac{1}{2}$. Since $z$ is an integer then $z \leq 28$.

When $z=28, x+y=57-28=29$.

We list all possible values for $x$ and $y$ in the table below given that $z=28$ and $x<y<z$.

| $y$ | 27 | 26 | 25 | 24 | 23 | 22 | 21 | 20 | 19 | 18 | 17 | 16 | 15 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $x$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |

Systematically, we continue decreasing the value of $z$ and listing all possible values for $x$ and $y$. When $z=27, x+y=57-27=30$.

| $y$ | 26 | 25 | 24 | 23 | 22 | 21 | 20 | 19 | 18 | 17 | 16 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $x$ | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |

When $z=26, x+y=57-26=31$.

| $y$ | 25 | 24 | 23 | 22 | 21 | 20 | 19 | 18 | 17 | 16 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $x$ | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |

When $z=25, x+y=57-25=32$.

| $y$ | 24 | 23 | 22 | 21 | 20 | 19 | 18 | 17 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $x$ | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |

When $z=24, x+y=57-24=33$.

| $y$ | 23 | 22 | 21 | 20 | 19 | 18 | 17 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $x$ | 10 | 11 | 12 | 13 | 14 | 15 | 16 |

When $z=23, x+y=57-23=34$.

| $y$ | 22 | 21 | 20 | 19 | 18 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| $x$ | 12 | 13 | 14 | 15 | 16 |

When $z=22, x+y=57-22=35$.

| $y$ | 21 | 20 | 19 | 18 |
| :--- | :--- | :--- | :--- | :--- |
| $x$ | 14 | 15 | 16 | 17 |

When $z=21, x+y=57-21=36$.

| $y$ | 20 | 19 |
| :--- | :--- | :--- |
| $x$ | 16 | 17 |

When $z=20, x+y=57-20=37$.

| $y$ | 19 |
| :--- | :--- |
| $x$ | 18 |

The next smallest value for $z$ is 19 and in this case $x+y=57-19=38$.

However, if $x+y=38$ then at least one of $x$ or $y$ must be 19 or larger.

This is not possible since $z=19$ and $x<y<z$.

Therefore, 20 is the smallest possible value for $z$ and we have listed the side lengths of all possible triangles above.

Counting, we see that there are $13+11+10+8+7+5+4+2+1=61$ possible triangles that satisfy the given conditions.

ANSWER: (B)

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