# $P$ is a point in $\triangle ABC$, with $\angle PBC=\angle PCB=24^\circ$, $\angle ABP=30^\circ$, $\angle ACP=54^\circ$. What is the $\angle BAP$?

Let $$P$$ be a point inside a triangle $$\triangle ABC$$ with $$\angle PBC=\angle PCB=24^\circ$$, $$\angle ABP=30^\circ$$ and $$\angle ACP=54^\circ$$. What is the $$\angle BAP$$?

I can calculate $$\angle BAP=18^\circ$$ by assuming $$BC=1$$ and calculating all the lengths involved in the picture using formula of sines and cosines.

But since the answer is an integer $$18^\circ$$, I believe there must be a more elementary ways (i.e. without using sines/cosines or invloving irrational numbers) to obtain the answers.

Any suggestions are welcome! Thank you in advance!

• Many similar questions have been asked before (for example this one) but not this exact one. Commented May 24, 2023 at 7:49
• Yes, a very nice Langley's Adventitious like configuration, one that i knew, not not exactly in this shape. A possible way to see the problem - by using the sine version of Ceva for the cevians through $P$, is (a posteriori)$$1=\frac{54^\circ}{24^\circ}\cdot\frac{24^\circ}{30^\circ}\cdot\frac{x=18^\circ}{48^\circ-x=30^\circ}\ .$$And the same relation holds if we permute the factors in the numerators, and/or those in the denominators. The configuration is knew was the one with the $24^\circ$ angles in same vertex... As as corollary, $$\sin18^\circ\cdot\sin54^\circ=\frac 14\ .$$Of course, +1... Commented May 27, 2023 at 0:01
• The above relation, product of the sines in $18^\circ$, $54^\circ$ is equivalent to$$\cos(54^\circ-18^\circ)-\cos(54^\circ+18^\circ)=\frac 12\ ,$$which brings us closer to the regular pentagon... Commented May 27, 2023 at 0:03
• Nice question! Where is it from?
– Dan
Commented May 27, 2023 at 13:27

1. Let $$Q$$ on $$AB$$ such that $$PQ \cong PB \cong PC$$.
2. By angle chasing, $$\measuredangle QPC = 108^\circ$$. Hence we can construct the regular pentagon $$PQRSC$$ shown above, where, in particular, $$AC$$ perpendicularly bisects $$QR$$.
3. $$\triangle AQR$$ is isosceles, and angle chasing yields $$\measuredangle AQR = 42^\circ$$.
4. Let now $$A'$$ on $$CA$$ (on the same side as $$A$$, with respect to $$PS$$) such that $$\triangle PA'S$$ is equilateral. We will show that $$A' \equiv A$$. We have in fact that $$\triangle A'QS$$ is isosceles, and angle chasing gives $$\measuredangle A'SQ = 24^\circ$$. Hence $$\measuredangle A'QS = 78^\circ$$, $$\measuredangle A'QR = 42^\circ$$, and by 3. we have our conclusion.
5. So $$\measuredangle PAS = 60^\circ$$, that is $$\measuredangle PAC = 30^\circ$$, hence the result $$\boxed{\measuredangle BAP = 18^\circ}.$$

Let draw regular pentagon BCDEF with center O. Let P is point such that $$\angle PBC=\angle PCB=24$$°, I is point such that $$\angle ICD=\angle IDC=24$$°, A$$_1$$ is point such that $$\angle A_1DE=\angle A_1ED=24$$°. Triangles PIC and A$$_1$$ID are regular and equal. Then triangle $$A_1IC$$ is isosceles with $$\angle A_1IC=360$$°$$-60$$°$$-132$$°$$=168$$°, then $$\angle ICA_1=6$$°, $$\angle PCA_1=54$$°.

A$$_1$$, O and B lie on perpendicular bisector of the edge DE, then $$\angle A_1BP=\angle OBP=30$$°. Then A$$_1$$ is the same point as A from problem statement. Then $$\angle BAP=\angle OA_1P=18$$°

• Thanks! I like your solutions the most! It’s short and explains the “hidden” symmetry of the original figure! Commented May 27, 2023 at 6:06

Here is a great party, so i will join. The point $$\Omega$$ below also shows up in many other solutions, but somehow i missed a quick argument.

Construct on $$BP$$ and $$CP$$ two equilateral triangles $$\Delta BPQ$$ and $$\Delta CPT$$ as in the picture. The perpendicular bisectors of $$PQ$$ (which also passes through $$A,B$$) and of $$PT$$ intersect in a point, call it $$\color{blue}\Omega$$, which also lies on the perpendicular bisector of $$BC$$. We obtain now by symmetries $$\color{blue}\Omega Q=\color{blue}\Omega P=\color{blue}\Omega T$$, and $$\Delta \color{blue}\Omega BC$$ isosceles, so we know the angles: $$\widehat{Q\color{blue}\Omega P}= \widehat{P\color{blue}\Omega T}=72^\circ\ .$$

This means that we can extend the broken line $$TPQ$$ to a regular pentagon $$TPQRS$$, with center of symmetry in $$\color{blue}\Omega$$ as in the picture. The marked angle in $$T$$ in the isosceles $$\Delta TSC$$ is $$60^\circ+108^\circ$$, so $$\widehat {TCS}=6^\circ$$, giving $$\widehat {SCP}=60^\circ-6^\circ=54^\circ=\widehat {ACP}$$. The point $$S$$ is thus $$S=B\color{blue}\Omega A \cap CA=A$$. Finally: $$x=\widehat{BSP}=\widehat{\color{blue}\Omega SP}= \frac 12\widehat{QSP}=\frac 12\cdot 36^\circ=\bbox[yellow]{\ 18^\circ \ }\ .$$

• Thanks! I believe the $BT$ in the forth line of your solution should be $PT$. Commented May 27, 2023 at 5:39
• It is interesting that the three main approaches here have in common that a point defined in some other way is shown to coincide with $A$. Personally, I tried to go the other way around and show directly that $A$ is the vertex of an equilateral triangle, but failed in doing so.
– dfnu
Commented May 27, 2023 at 11:11
• Yes, it is $PT$ at an important first point in the story, correcting it, thanks! Commented May 27, 2023 at 13:27

I'm a 9th grader in Viet Nam so my English won't be too good. Sorry for this inconvenience.

My Solution:

Let $$\angle BEP =E_1,\angle PEH = E_2,\angle AEH = E_3,\angle FEA = E_4$$

I will draw $$AG\bot BC \text{ , }PD\bot BC \text{,}AF\bot AG,FD\cap AB={E},EH\text{ bisector }\angle PEA$$so I will have $$FD||AG\to\angle BED=\angle BAG\text{ (Corresponding pair)}$$ We can easily calculate: $$\angle BED\text{ in }\triangle BED,\measuredangle BED=180^o-\angle EDB-(\measuredangle EBP+\measuredangle PBD)=180^o-90^o(30^o+24^o)=36^o(1)$$ But we have: $$\angle BED=\angle BAG$$(Proved above) $$\Rightarrow \measuredangle BAG =36^o$$ and $$EH$$ bisector $$\angle PEA\to \angle PEH = \angle AEH$$ but we also have $$\angle BEP = \angle FEA = 36^o(\text{opposite angles})\Rightarrow \angle BEP+\angle PEA = 180^o \text{and}\angle FEA + \angle AEP =180^o(\text{both are complementary angle})$$ $$\to36^o + \angle PEA=180^o \text{and }36^o+\angle AEP = 180^o$$ or $$E_2+E_3=144^o \text{ we also have EH bisector }\angle PEA\to E_2=E_3=72^o$$ .On the other hand $$EP||AI \text{ because }FD||AG\to\angle PEI =\angle EIA(\text{Alternate interior})\to\angle AEH = \angle AIH = 72^o\to \triangle AEI \text{ is a Isosceles triangle}$$

Draw $$AJ\bot EI\to AJ \text{ is also bisector }\angle EAI\to \angle EAJ = \frac{1}{2}\angle EAI =\frac{1}{2}\angle BAG=18^o$$ $$\Rightarrow EAP = 18^o$$

• Why AJ passes through P? Commented May 24, 2023 at 23:42
• @LydiaLee You're right, I should have read this more carefully. Indeed, the angle $BAJ$ equals $18$, but that's not equal to angle $BAP$. Indeed, one precisely needs to prove that $P$ lies on $AJ$ for this to occur. That is equivalent to proving that $EH$ is perpendicular to $AP$, which is shown but not proved by the author. Commented May 25, 2023 at 11:09
• And , in fact, that cannot be proved without alluding to the angles on the other side i.e. $24$ and $54$, which have not found any use yet. Surely they are important for this problem. Commented May 25, 2023 at 11:12
• Finally, in the edited "written" part of this answer, angle JPD is claimed to be equal to 18, but that uses the fact that AJ passes through P, which was to be proved. Commented May 25, 2023 at 11:23
• Has this incomplete solution ever used that $\angle ACP=54$° ? Commented May 25, 2023 at 11:41

We may conveniently partition $$\angle BAC$$ as $$\angle BAP=18^\circ+\alpha$$ and $$\angle CAP=30^\circ-\alpha$$, where $$-18^\circ<\alpha<30^\circ$$. Comparing the triangles $$BAP$$ and $$CAP$$, we note that $$AP$$ is common and $$|BP|=|CP|$$ (since $$\triangle BPC$$ is isosceles). Hence, applying the sine rule in a similar way to each triangle yields $$\sin30^\circ\sin(30^\circ-\alpha)=\sin54^\circ\sin(18^\circ+\alpha)$$, which may be written $$\cos60^\circ\cos(60^\circ+\alpha)=\cos36^\circ\cos(72^\circ-\alpha).\qquad\qquad(1)$$ It is well known that $$\cos36^\circ=\frac14(\surd5+1)$$ and $$\cos72^\circ=\frac14(\surd5-1)$$. (A proof is appended.) Hence $$\cos36^\circ\cos72^\circ=\frac14=\cos^260^\circ$$, and therefore $$\alpha=0$$ is a solution to eqn $$1$$. Also, the LHS of eqn $$1$$ is decreasing while the RHS is increasing in the given range for $$\alpha$$. Hence the solution $$\alpha=0$$ is unique.

Appendix$$\quad$$ A property of the cosine is that $$\cos(3\times72^\circ)=-\cos36^\circ=\cos(2\times72^\circ).$$ From the identities $$\cos2\theta=2\cos^2\theta-1$$ and $$\cos3\theta=4\cos^3\theta-3\cos\theta$$, it follows that $$x=\cos72^\circ$$ is a solution of the cubic $$4x^3-3x=2x^2-1$$, or $$(x-1)(4x^2+2x-1)=0.$$ Since $$0<\cos72^\circ<1$$, the linear factor may be ignored, and the required value is the positive root of the quadratic: $$\cos72^\circ=\frac14(\surd5-1)$$. The value $$\cos36^\circ=\frac14(\surd5+1)$$ follows from the initially mentioned properties.

• I believe author of original post knows about this way of solution and seeks for another way Commented May 26, 2023 at 20:26
• @IvanKaznacheyeu : The OP said that he calculated $18^\circ$, and indeed that was the answer my calculator gave me when I first looked at the problem. However, I found it less easy to prove that the angle was exactly $18^\circ$. Although he didn't ask for a trigonometric solution, and he may have found this or a better one, I think that it might be useful to post such a solution, properly done, for the sake of comparison. I didn't find any of the other answers easy to follow. Commented May 26, 2023 at 21:06
• OP also wrote about irrational numbers, I believe OP knows this (or similar) way of solution Commented May 26, 2023 at 21:13
• It is not difficult to obtain the exact values of $sin$/$\cos$ of all the angles involved in questions ($18^\circ,36^\circ,54^\circ$ etc.) So I think conceptually it is not difficult to verify $18^\circ$ is the answer. Thank you anyway! Commented May 27, 2023 at 5:53

Construct a regular pentagon on $$BC$$. Since $$\angle ABC=54^o$$, then $$BA$$ bisects $$\angle FBC=108^o$$, and $$BAK$$ is the perpendicular bisector of $$GH$$. Similarly, since $$\triangle BPC$$ is isosceles, $$GPJ$$, crossing $$BK$$ at $$L$$, perpendicularly bisects $$BC$$, and $$FLM$$ perpendicularly bisects $$CH$$.

Thus if we join $$PH$$, $$AG$$, and $$AH$$, we have $$\triangle PGH\cong\triangle ABC$$, and $$\triangle HAG\cong\triangle BPC$$.

By symmetry then, $$\triangle LAP$$ is isosceles. And since exterior $$\angle KLG=180^o-90^o-54^o=36^o$$, then $$\angle LAP=\angle APL=18^o$$.

• How do you know that $\triangle PGH\cong\triangle ABC$ ? Commented May 26, 2023 at 7:24
• @Piita: the two mentioned triangles have a congruent side, and the opposite vertex of each of them lies on the other triangle perpendicular bisector. But without knowlegde of angles one cannot conclude that the triangles are congruent, I believe. (Basically, the "position" of $A$ and $P$ on each perpendicular bisector is not shown to be the same.)
– dfnu
Commented May 26, 2023 at 13:30
• @dfnu I stand corrected, there's an angle-chasing part missing, to show AK=PJ, or that AP is indeed perp. to LM or parallel to CH. Commented May 26, 2023 at 15:56
• @IvanKaznacheyeu—Your question and the other comments are all justified; my argument is flawed. Commented May 26, 2023 at 20:25