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In the diagram below, $ABC$ is a triangle. Given that $\overline{AD}=\overline{BC}$, $\angle ABC=120^{\circ}$, $\angle BDA=3\phi$, and $\angle BCA=2\phi$, determine the measure of $\phi$.

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Construct the equilateral triangle $BQC$ and the parallelogram $ABPD$. Angle chasing gives $\angle DBC = \phi$, $\angle DAB\cong PBQ \cong BPD=60^{\circ}-2\phi$, $\angle BDP = 120^{\circ}-\phi$. $\triangle BPQ$ is isosceles, thus $\angle BQP\cong\angle BPQ=60^{\circ}+\phi$. It follows that $BQPD$ is cyclic. It follows that $\angle BPD=60^{\circ}-2\phi$, thus $\angle BQD=60^{\circ}-2\phi$. It follows that $\triangle BQD$ is isosceles, and so is $\triangle DQC$. It follows that $\angle DBQ\cong\angle QDB = 60^{\circ}+\phi$. It follows that the angles of the triangle $DQC$ is $60^{\circ}+2\phi+60^{\circ}+2\phi+2\phi$. Thus $\phi=10^{\circ}$.


It feels as if there should be a simpler geometric solution. Can you come up with one?

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  • $\begingroup$ @Piquito Are you saying either of those conditions is enough to solve the problem? How? $\endgroup$ – blackened May 12 at 6:10
  • $\begingroup$ I think I made a mistake. I'm sorry $\endgroup$ – Piquito May 13 at 16:46
  • $\begingroup$ Assuming $\overline{AD}=\overline{BC}$ one has the relation between $\theta=\angle ABC,\alpha=\angle BCA$ and $\beta=\angle BDA$ $$\frac{\sin(\alpha+\theta)}{\sin(\theta)}=\frac{\sin(\beta)}{\sin(\beta)+\sin(\beta-\alpha)}$$ and if we take $2\alpha$ and $3\alpha$ instead of $\alpha$ and $\beta$ respectively we get $$\frac{\sin(2\alpha+\theta)}{\sin(\theta)}=\frac{\sin(3\alpha)}{\sin(3\alpha)+\sin(\alpha)}$$ Consequently a value of $\theta$ in the ABC triangle corresponds to a given value of $\alpha $. Your problem was built knowing in advance the two values of $10$ and $120$ degrees. $\endgroup$ – Piquito May 13 at 17:14
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[I like your synthetic solution. One could always brute force it out with trigo.]

From sine rule on ABD, $ \frac{ BA} { \sin 3 \phi} = \frac{AD}{ \sin (120^\circ - \phi)}$

From sine rule on BAC, $\frac{BA}{ \sin 2 \phi} = \frac{BC} { \sin (60 ^ \circ - 2 \phi)}$

Hence, $\frac{ \sin 3 \phi} { \sin (120^\circ - \phi) } = \frac{BA}{AD} = \frac{BA}{BC} = \frac { \sin 2 \phi} { \sin ( 60 ^ \circ - 2 \phi )} $

Expanding this and solving for $\phi$, bearing in mind that $\phi < 30 ^ \circ $ from geometric considerations, gives us $\phi = 10 ^ \circ$.

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    $\begingroup$ A GEOMETRIC solution is required by the O.P. $\endgroup$ – Piquito May 10 at 15:59
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    $\begingroup$ Trigonometry is geometry, but I'd let OP decide. If he intends for a "synthetic geometry" approach, I would be somewhat surprised if there's one that's significantly better than his. $\endgroup$ – Calvin Lin May 10 at 16:01
  • $\begingroup$ @Calvin Why would you be surprised? I've seen countless tough-looking geometry problems which were solved with a brilliant idea almost in a single (explanatory) line (along with an accompanying diagram). $\endgroup$ – blackened May 10 at 16:09
  • $\begingroup$ Convention is that geometric $\ne$ trigonometric. Regards. $\endgroup$ – Piquito May 10 at 16:09
  • $\begingroup$ @Blackened Mainly because the construction of Q is extremely motivated by the setup, and that somewhat requires P to push through any of the following details. I tried working with Q without P, but wasn't able to show (say) that QB, QD, QC had the same length. $\endgroup$ – Calvin Lin May 10 at 16:13
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Construct a point $D'$ inside $\triangle ADB$, such that $\triangle DCB\cong \triangle D'AD$, which is possible since $\vert AD\vert=\vert CB\vert$. Now reflect $D'$ along $AD$ in order to obtain $D''$ which, again, leads to $\triangle ADD''\cong \triangle ADD'\cong\triangle DCB$.

Observe now that $$\triangle D''DD'\cong \triangle D'DB\implies \color{green}{2\cdot \vert D'E\vert=\vert D'B\vert}\tag{1}$$ Angle chasing in $\triangle ACB$ shows $$\angle CAB=60°-2\phi\implies \angle D'AB=60°-2\phi-\angle DAD'=60°-4\phi\tag{2}$$ Thus $\angle D''AB=60°$. Furthermore, angle chasing confirms that $$\angle DBC=\angle BDA-\angle BCD=\phi\qquad \angle D'BD=\frac{180°-\angle BDD'}{2}=\frac{180°-2\phi}2=90-\phi$$ Putting this together yields $$\angle ABD'=120°-\angle D'BD-\angle DBC=30°$$ If we extend $BD'$ to intersect $AD''$ at $F$, we notice that $\triangle AFB$ is a $30°-60°-90°$ triangle.

Let $G$ be the point on $AB$ such that $\angle D'GB=90°$. Now, $\triangle BGD'\sim \triangle AFB$. Hence $$\color{green}{2\cdot \vert GD'\vert=\vert D'B\vert\stackrel{(1)}{=}2\cdot\vert D'E\vert\implies \vert GD'\vert=\vert D'E\vert}$$ And we are almost done, since this implies that $AD'$ bisects $\angle DAB$. Therefore, and finally $$2\phi=\angle DAD'=D'AB\stackrel{(2)}{=}60°-4\phi\iff 2\phi=60°-4\phi\iff\fbox{$\ \phi=10°\ $}$$


Sidenote $\ $ I don't know if this solution is simpler than the one you've offered. What I do know, is that this proof is almost elementary after one has introduced all points I use and has performed angle chasing...

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  • $\begingroup$ If you are forbidden to use trigonometry, this problem is not trivial, believe me (at least for the fact of having put you an upvote) You can verify what I say giving this problem to your friends knowing maths. $\endgroup$ – Piquito May 11 at 20:05
  • $\begingroup$ @Dr. The problem with “... this proof is almost trivial after one has introduced all points I use ...” is that in many (if not most) geometry solutions, once the auxiliary construction(s) are made, the rest follows trivially. I would think, coming up with that point D trick is anything but trivial. $\endgroup$ – blackened May 12 at 14:35

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