Question 15·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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A convex polygon has interior angles (in degrees) that form an arithmetic sequence. The smallest interior angle measures $120^\circ$ , and each interior angle is $5^\circ$ greater than the previous one.

Which choice is the number of sides of the polygon?

Angles that rise by a fixed amount form an arithmetic sequence, so add them using the average of the first and last angles. A polygon with $n$ sides has $(n-2)$ triangles, giving its interior-angle total. Set the two totals equal and solve with a Desmos regression. Convex matters: a solution that makes any interior angle $180^\circ$ or more cannot describe the polygon.

Hints

Hint 1
In an arithmetic sequence, each term increases by the same amount. With $n$ angles, there are $n-1$ increases from the first angle to the last. How would you write the last angle?
Hint 2
A convex polygon has every interior angle less than $180^\circ$ . How many $5^\circ$ increases from $120^\circ$ reach $180^\circ$ , and what angle number is that?
Hint 3
Add an arithmetic sequence by multiplying its first-and-last average by the number of terms. A polygon’s angles also total $180(n-2)$ because it splits into $n-2$ triangles. Which totals must match?

Step-by-step

Match the sequence sum to the polygon angle sum
Step 1
Write the last angle
Let $n$ be the number of sides, so there are $n$ interior angles. Starting at $120^\circ$ , you make one fewer increase than the number of angles to reach the last one. Its measure is $120+5(n-1)$ degrees.
Step 2
Limit the number of sides
Type $(180-120)/5$ into Desmos to find how many $5^\circ$ increases reach $180^\circ$ ; it shows $12$ . Starting at the first angle, 12 increases reach the 13th angle. Every angle of a convex polygon is less than $180^\circ$ , so $n<13$ . Keep this limit when solving; the angle-sum equation alone may allow an invalid result.
Step 3
Add the angle sequence
The arithmetic-sequence sum is the number of angles times the average of the first and last angles. Using the last angle you found, the total is
$\frac{n}{2}\bigl(120+[120+5(n-1)]\bigr).$
Step 4
Match the polygon’s angle total
Drawing diagonals from one corner splits an $n$ -sided polygon into $n-2$ triangles. Each contributes $180^\circ$ , so the sequence sum must equal the triangle-based polygon sum:
$\frac{n}{2}\bigl(120+[120+5(n-1)]\bigr)=180(n-2).$
Step 5
Solve for a valid number of sides
Type the equation in Desmos with $=$ changed to $\sim$ so Desmos fits $n$ . Add $\{3\le n<13\}$ to keep the result within the valid range: a polygon has at least $3$ sides, and convexity gave $n<13$ . Desmos reports $n=9$ under PARAMETERS. So the polygon has 9 sides. Choice C.

Question 15·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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