Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME Prove it United States

Problem:

Let RSTUVR S T U V be a regular pentagon. Construct an equilateral triangle PRSP R S with point PP inside the pentagon. Find the measure (in degrees) of angle PTVP T V.

Solution

Solution:

We have ∠PRV=∠SRV−∠SRP=108∘−60∘=48∘\angle P R V=\angle S R V-\angle S R P=108^{\circ}-60^{\circ}=48^{\circ}. Since PR=RS=RVP R=R S=R V, triangle PRVP R V is isosceles, so that ∠VPR=∠RVP=(180∘−∠PRV)/2=66∘\angle V P R=\angle R V P=\left(180^{\circ}-\angle P R V\right) / 2=66^{\circ}. Likewise, we have ∠TPS=66∘\angle T P S=66^{\circ}, so that
∠TPV=360∘−(∠VPR+∠RPS+∠SPT)=360∘−(66∘+60∘+66∘)=168∘ \angle T P V=360^{\circ}-(\angle V P R+\angle R P S+\angle S P T)=360^{\circ}-\left(66^{\circ}+60^{\circ}+66^{\circ}\right)=168^{\circ}
Finally, by symmetry, triangle PTVP T V is isosceles (PT=TV)(P T=T V), so ∠PTV=∠TVP=(180∘−∠TPV)/2=6∘\angle P T V=\angle T V P=\left(180^{\circ}-\angle T P V\right) / 2=6^{\circ}. (See the figure.)

Figure 1

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.