Competition · AMC preparation · step 4 of 4
AMC 8 · 1999 · #21
Grade 7 geometry-2d
Pick an answer.
AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The figure looks tangled, but the three marked angles each sit inside a small triangle formed by the crossing segments. Tool #7 (Identify Subproblems) breaks the picture into just two of those triangles: one that holds the 100° and 40° marks, and a second one that holds the 110° mark together with angle A. The two triangles share a side, so the third angle of the first triangle reappears as a known angle in the second. Tool #1 (Draw a Diagram) is the bookkeeping partner — mark each angle on the figure as you find it so the transfer between the two triangles is visible. No algebra needed; the triangle-angle-sum rule does all the work.
Find the third angle
First subproblem: the triangle holding 100° and 40°. Its angles sum to 180°, so the third angle is 40°.
Once two angles of a triangle are known, the third is just 180° minus their sum.
7.G.B.5Identify SubproblemsCarry the angle to the next triangle
That 40° sits on the side shared with the next triangle, so it is also an interior angle of the triangle holding angle A and 110°.
The two triangles share a side, so the angle that sits along that shared side is the same in both triangles. Drawing it on the figure makes the reuse obvious.
4.G.A.1Draw A DiagramUse the angle sum again
Close the second triangle with the angle-sum rule: A, 110°, and the transferred 40° give A = 30°.
The same rule that closed the first triangle closes the second — two passes of "angles add to 180°" is the whole proof.
Angle A is the third corner of a triangle whose other two corners measure 110° and 40°, so its size is 180° minus those two.
▸ Why?
The three corners of that triangle — angle A, the 110° mark, and a 40° angle — must total 180°, so once the 110° and 40° are fixed, angle A is forced to be the leftover.
▸ Why?
The 40° used here is not marked on the picture; it is carried over from the neighboring triangle that holds the 100° and 40° marks.
▸ Why?
That neighboring triangle already shows two of its corners, 100° and 40°, so its third corner has to be 180° minus 100° minus 40°, which is 40°.
▸ Why?
The A-triangle's corner at the crossing point sits directly opposite that 40° corner where the two straight segments cross, and opposite angles there are equal because each one fills the same straight segment together with the shared neighbor angle, so this corner is also 40°.
Two small triangles, one shared side. The first triangle pins down a 40° angle (180 - 100 - 40); reusing it in the second triangle forces A = 180 - 110 - 40 = 30° — choice (B).
- Find the third angle
- Carry the angle to the next triangle
- Use the angle sum again
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