AMC 8 · 2000 · #13
Grade 8 geometry-2dProblem
In triangle , we have and . If bisects , then
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Triangle $CAT$ is isosceles with $\angle ACT = \angle ATC$ and apex angle $\angle CAT = 36^\circ$. Segment $\overline{TR}$ bisects $\angle ATC$, with $R$ on side $\overline{CA}$. Find $\angle CRT$.
Givens: $\triangle CAT$ has $\angle ACT = \angle ATC$ (so it is isosceles with apex at $A$); $\angle CAT = 36^\circ$; $\overline{TR}$ bisects $\angle ATC$, with $R$ on side $\overline{CA}$; Answer choices: (A) $36^\circ$, (B) $54^\circ$, (C) $72^\circ$, (D) $90^\circ$, (E) $108^\circ$
Unknowns: The measure of $\angle CRT$
Understand
Restated: Triangle $CAT$ is isosceles with $\angle ACT = \angle ATC$ and apex angle $\angle CAT = 36^\circ$. Segment $\overline{TR}$ bisects $\angle ATC$, with $R$ on side $\overline{CA}$. Find $\angle CRT$.
Givens: $\triangle CAT$ has $\angle ACT = \angle ATC$ (so it is isosceles with apex at $A$); $\angle CAT = 36^\circ$; $\overline{TR}$ bisects $\angle ATC$, with $R$ on side $\overline{CA}$; Answer choices: (A) $36^\circ$, (B) $54^\circ$, (C) $72^\circ$, (D) $90^\circ$, (E) $108^\circ$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Break Into Subproblems
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 $\overline{TR}$ cuts the big triangle $\triangle CAT$ into a small triangle $\triangle CRT$ that contains the angle we want. Tool #7 (Break Into Subproblems) splits the work into two short angle-sum steps: first use $\triangle CAT$ to find the base angle $\angle ATC$, then use $\triangle CRT$ (whose two other angles we now know) to find $\angle CRT$. No algebra needed beyond "angles in a triangle add to $180^\circ$."
Execute — Answer: C
8.G.A.5 Step 1 - Read the big triangle.
- $\triangle CAT$ has $\angle ACT = \angle ATC$ (the two base angles) and apex angle $\angle CAT = 36^\circ$.
- The three angles add to $180^\circ$, so the two equal base angles together account for $180^\circ - 36^\circ = 144^\circ$.
💡 Grade 8 "angle sum of a triangle is $180^\circ$"; the two equal base angles split the leftover $144^\circ$ evenly.
7.G.B.5 Step 2 - Apply the bisector.
- $\overline{TR}$ cuts $\angle ATC = 72^\circ$ in half, so the piece inside the small triangle $\triangle CRT$ is $\angle RTC = 36^\circ$.
- Mark $36^\circ$ at $T$ on the picture.
💡 Grade 7 angle facts: a bisector divides an angle into two equal halves.
8.G.A.5 Step 3 - Read the small triangle.
- Inside $\triangle CRT$ the corner at $C$ is the same $\angle ACT = 72^\circ$ (point $R$ lies on side $\overline{CA}$, so the angle at $C$ is unchanged), and the corner at $T$ is the bisected piece $\angle RTC = 36^\circ$.
- The three angles must add to $180^\circ$.
💡 Same angle-sum law applied to the smaller triangle, with the two known angles subtracted from $180^\circ$.
8.G.A.5 Read the big triangle. $\triangle CAT$ has $\angle ACT = \angle ATC$ (the two ba 7.G.B.5 Apply the bisector. $\overline{TR}$ cuts $\angle ATC = 72^\circ$ in half, so the 8.G.A.5 Read the small triangle. Inside $\triangle CRT$ the corner at $C$ is the same $\ Review
Reasonableness: Two sanity checks. (i) Angle sum in $\triangle CRT$: $72^\circ + 36^\circ + 72^\circ = 180^\circ$. (ii) The exterior-angle view: $\angle CRT$ is the exterior angle of $\triangle ART$ at $R$, so it equals the sum of the two remote interior angles, $\angle RAT + \angle ATR = 36^\circ + 36^\circ = 72^\circ$ — same answer. The trap choices line up with common slips: (A) $36^\circ$ stops at the apex angle or the bisected half; (B) $54^\circ$ comes from halving $108^\circ$ as if the apex were the base angle; (E) $108^\circ$ is the supplement $180^\circ - 72^\circ$, i.e. the other angle at $R$ on segment $\overline{CA}$.
Alternative: Tool #10 (Use a Related Problem): this is the classic "golden gnomon" $36\text{-}72\text{-}72$ isosceles triangle. Bisecting one base angle is the standard construction that splits it into two smaller isosceles triangles. The bottom piece $\triangle CRT$ has angles $72^\circ$ at $C$, $36^\circ$ at $T$, hence $72^\circ$ at $R$ — automatically isosceles with $CR = CT$ — giving the answer (C) without arithmetic.
CCSS standards used (min grade 8)
8.G.A.5Use informal arguments about angle sums of triangles (Applying "the three angles of a triangle sum to $180^\circ$" twice: first to $\triangle CAT$ to get $\angle ATC = 72^\circ$, then to $\triangle CRT$ to get $\angle CRT = 72^\circ$.)7.G.B.5Use facts about supplementary, complementary, vertical, and adjacent angles to solve simple equations (Reading the bisector relation $\angle RTC = \tfrac{1}{2}\angle ATC = 36^\circ$ and recognizing $\angle RCT$ is the same as $\angle ACT$ because $R$ lies on side $\overline{CA}$.)
⭐ Label every angle you can on the picture, then the small triangle $\triangle CRT$ has angles $72^\circ$ and $36^\circ$ already known — the third angle has to be $180^\circ - 72^\circ - 36^\circ = 72^\circ$.
⭐ Label every angle you can on the picture, then the small triangle $\triangle CRT$ has angles $72^\circ$ and $36^\circ$ already known — the third angle has to be $180^\circ - 72^\circ - 36^\circ = 72^\circ$.
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