AMC 8 · 2000 · #13
Grade 8 geometry-2d
Pick an answer.
AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The figure is given, so Tool #1 (Draw a Diagram) tells us to read every angle right off the picture rather than set up equations. The bisector TR cuts the big triangle △ CAT into a small triangle △ CRT that contains the angle we want. Tool #7 (Break Into Subproblems) splits the work into two short angle-sum steps: first use △ CAT to find the base angle ∠ ATC, then use △ CRT (whose two other angles we now know) to find ∠ CRT. No algebra needed beyond "angles in a triangle add to 180°."
Angle sum in △ CAT: the two equal base angles split 180° - 36°, so each base angle ∠ ATC = 72°.
Grade 8 "angle sum of a triangle is 180°"; the two equal base angles split the leftover 144° evenly.
8.G.A.5Draw A DiagramTR bisects ∠ ATC = 72°, so the half inside △ CRT is ∠ RTC = 36°.
Grade 7 angle facts: a bisector divides an angle into two equal halves.
7.G.B.5Identify SubproblemsIn △ CRT the angle at C stays ∠ ACT = 72° and at T is ∠ RTC = 36°, so ∠ CRT = 72°.
Same angle-sum law applied to the smaller triangle, with the two known angles subtracted from 180°.
8.G.A.5Draw A DiagramLabel every angle you can on the picture, then the small triangle △ CRT has angles 72° and 36° already known — the third angle has to be 180° - 72° - 36° = 72°.