Question on the regular pentagon and triangle

Question

CONTEXT

I was recently working on some Euclidean geometry problems that involved regular pentagons and triangles, such as this one, and this one (to which the current question specifically refers). I noted that sometimes I was "forced" to introduce a "dummy" point that I then proved being coincident with the point actually involved in the hypotheses. My quesiton is, therefore, if this procedure is really needed, and, if so, "why".

THE PROBLEM

Here is the problem statement, that, as mentioned, is part of this proof.

Let $PQRST$ be a regular pentagon, and let $U$ be a point on the perpendicular bisector of $PQ$, such that $\measuredangle PQU = 42^\circ$. Prove that $\triangle RTU$ is equilateral.

MY APPROACH

I solved the problem by introducing the equilateral triangle $\triangle RTU'$. Angle chasing and the fact that $\triangle TQU'$ is isosceles leads to the required result, that is $\measuredangle PQU' = 42^\circ$, and therefore $U'\equiv U$.

MY QUESTION

I would say I formally showed the reversed implication, i.e. that if the triangle is equilateral than $\measuredangle PQU = 42^\circ$. However, since there is only one point $U$ on the perpendicular bisector of $PQ$ with the property $\measuredangle PQU = 42^\circ$, then we can also state that the original assertion is true.

Is there a more direct approach, not involving the point $U'$? If not, does it make any sense to ask oneself why this is not possible? And, if so, what is the answer to this question?

Answer 1

Let $\alpha = \angle PQU$.

We want to show that $\alpha = \frac{7 \pi}{30} = 42°$.

The way I have obtained the result is by applying the sine law to triangle $RQU$. Here is how.

• $\angle URQ = \angle SRQ -(\angle SRT + \angle TRU)=\frac{3 \pi}{5}-\frac{\pi}{5}-\frac{\pi}{3}=\frac{\pi}{15}.$

• $\angle RQU =\frac{3 \pi}{5}+\alpha.$

• as a consequence (sum of angles = $\pi$) :

$$\angle QUR =\frac{\pi}{3}-\alpha \tag{0}$$

Please note that (0) implies that $\alpha \in [0,\tfrac{\pi}{3}]$.

Besides,

• $RQ = 2 \sin \frac{\pi}{5}$ (pentagon's side),

• $UR = 2 \sin \frac{2\pi}{5}=4 \sin\frac{\pi}{5}\cos\frac{\pi}{5}$ (pentagon's diagonal)

We are now able to apply sine law in triangle $PQU$ :

$$\frac{\sin\left(\frac{3 \pi}{5}+\alpha\right)}{UR}=\frac{\sin\left(\frac{\pi}{3}-\alpha\right)}{RQ} \ \iff \ \frac{\sin\left(\frac{3 \pi}{5}+\alpha\right)}{4 \sin\frac{\pi}{5}\cos\frac{\pi}{5}}=\frac{\sin\left(\frac{\pi}{3}-\alpha\right)}{2 \sin\frac{\pi}{5}}$$

$$\sin\left(\frac{3 \pi}{5}+\alpha\right)=2\cos\frac{\pi}{5}\sin\left(\frac{\pi}{3}-\alpha\right)\tag{1}$$

In the domain of possible variation $[0,\tfrac{\pi}{3}]$ for $\alpha$, the LHS of (1) is decreasing (negative derivative) whereas its RHS is increasing ; therefore if there is a solution $\alpha$ to (1) (and we know there is one !), it is unique.

It suffices to check that taking $\alpha=\frac{7\pi}{30}$ gives an identity in (1). A small calculation shows that it amounts to verify that :

$$\underbrace{\sin\frac{5\pi}{6}}_{= \ 1/2}=2 \ \underbrace{\cos\frac{\pi}{5}}_{\Phi/2}\underbrace{\sin\frac{\pi}{10}}_{\tfrac12 (\Phi-1)}$$

(using this answer and formula $\sin^2a=\frac12(1-\cos(2a))$, where $\Phi$ is the golden ratio), with an immediate verification (due to relationship $\Phi^2=\Phi+1$).

Answer 2

I finally managed to find a direct "synthetic" approach.

Recall that our hypothesis is that $PQRST$ is a regular pentagon, $U$ lies on the perpendicular bisector of $PQ$, with $\measuredangle PQU = 42^\circ$. We want to show that $\triangle RTU$ is equilateral.

Since $\triangle RTU$ is isosceles, we need to show that $\measuredangle RTU = 60^\circ$.

Construct the equilateral triangles $\triangle STW$ and $\triangle QUV$ as shown above.

1. Angle chasing shows that $\measuredangle WPQ = 42^\circ$ (so that $PW \parallel QU$) and $\measuredangle WQU = \measuredangle QUP = 96^\circ$. This means that $PUQW$ is an isosceles trapezoid, and, in particular $PU \cong QW$.
2. This, together with the fact that $\measuredangle VQW = \measuredangle VUP = 36^\circ$, yields $\triangle VQW \cong \triangle VUP$ (SAS).
3. Therefore, $\triangle PVW$ is isosceles.
4. $TPVW$ is a kite, and, thus $\measuredangle PTV = \measuredangle VTW = 24^\circ$.
5. Since by construction and angle chasing $\measuredangle TPV = 78^\circ$, we have (thanks to 4.) that $\triangle TPV$ is isosceles.
6. $TPUV$ is then also a kite, and $\measuredangle PTU = \measuredangle UTV = 12^\circ$.
7. Thus $\measuredangle RTU = 60^\circ$, and we reach our thesis.