AMC 8 1999 Problem 21: Geometry Solution

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Answer: B — 30°.

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Hint 1 of 2

The 100° and 110° marks sit at crossing points, so each has a partner angle: its supplement (straight line) and its vertical angle (the X across the crossing). Convert the marked angles into the angles that actually live inside the small triangles.

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Hint 2 of 2

The cleanest tool here is the exterior-angle theorem: in any triangle, an exterior angle equals the sum of the two non-adjacent interior angles. Chase from the known tips toward A.

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Approach: exterior-angle theorem to chase toward A

First turn the marks into triangle angles. The 100° has supplement 80° on the line through A; the 110° has supplement 70°.
Look at the triangle with the 40° tip and the 70° angle: its third angle is 180° − 40° − 70° = 70°. The vertical angle at that same crossing (on A's triangle) is also 70°.
Now A's triangle has the 80° and that 70°: ∠A = 180° − 80° − 70° = 30°.
Tools you'll reuse on every star/chase: at a crossing, the supplement (sums to 180° along a line) and the vertical angle (equal across the X) let you teleport a known angle into a neighboring triangle — then the 180° triangle sum finishes it.

Another way — the five star tips sum to 180°:

A famous fact: the five point-angles of a 5-pointed star always add to 180°. Two of the tips here can be read from the figure's marked angles.
Using the supplements, the tip angles work out so that the remaining tip A fills the gap to 180°, giving ∠A = 30°.
Knowing the 'star tips = 180°' result lets you skip most of the chase once you can identify the other tip angles — a handy shortcut to verify the answer.