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Let $\triangle ABD$ be a triangle and $C$ be a point on $BD$ such that $ {AB} ={CD}$, $\angle BAC = 18^\circ$ and $\angle ABC = 12^\circ$. The question is to find $x$, where $\angle ADC = x^\circ$.

I think this is a pretty classic problem but I am not able to find a solution using the techniques which were used to solve a similar kind of problem (only the angles were different and they were $30^\circ$ and $24^\circ$ respectively). The idea was to create an equilateral triangle by drawing a perpendicular line segment from $B$ to $AC$ with the same length as $AB$. However, that technique does not work here and I cannot find another way. Any help is appreciated (I forgot to add this but an answer which uses "geometrical" methods like drawing line segments or forming new triangles etc. would be great), thank you. The figure is as follows:

triangle

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    $\begingroup$ Can you figure out what $x$ is via an accurate diagram? That might give you some ideas for other things to try. $\endgroup$ – Calvin Lin Nov 17 '20 at 21:49
  • $\begingroup$ @CalvinLin I used the Law of Sines and a calculator to get an exact value of $x$. Then I spent 2 hours trying for a geometric demonstration, per the OP's request. No joy. If you have a hint re auxiliary lines to draw, I (for one) would really like to know it. $\endgroup$ – user2661923 Nov 18 '20 at 0:53
  • $\begingroup$ @user2661923 The law of sines is how I would have approached this problem. I suspect that there is a "geometric" solution, but nothing springs to mind as yet. $\endgroup$ – Calvin Lin Nov 18 '20 at 1:19
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I will insert a slow geometric solution. Doing it in steps may allow a pause after any step, the ambitious reader may want to find her or his own path to the solution. This may be also a good preparation for an olympiad or related challenge... The reader in hurry may chose to consume one of the two solutions listed in (4) and/or (5), which can be understood independently.


(1) Some picture:

In such situations it may be useful to use the pattern of an existing symmetry inside a regular polygon. The given triangle $\Delta ABC$ should be realized with vertices among the vertices of the polygon. (But the imagination of the solver should not stop here, sometimes using the center or some related intersection of diagonals may be also fruitful.) In our case we search a constellation with $A,B,C$ among the vertices of the regular polygon $P_{30}$. If the vertices are $0,1,2,3,\dots,14, 15=-15, 16=-14,\dots, 29=-1$, denoted using the elements of $\Bbb Z/30$, then a possible choice is $B=0$, $C=3$, $A=5$.

We draw first a circle $(O)$ centered in some point $O$. Since the angles with measures $12^\circ$ and $18^\circ$ are given, we expect to "see" also an inscribed angle in $(O)$ with measure $6^\circ$, so we decide to draw a $30$-gon inscribed in $(O)$. We fix the position of $A$, make then the one or the other choice of the position of $B$ and obtain the following picture:

mathematics stackexchange picture for problem 3911812, fig1

In the figure, we have also constructed the point $D$ as follows. Let $(A)$ be the circle centered in $A$ with radius $AB$. Let $S,T$ be the two points of intersection of $(A)$ with the parallel trough $A$ to $BC$. Here $S$ is the point making $\Delta ABS$ to have angles $<90^\circ$, so the angles are $12^\circ$ in $A$, and $84^\circ$ in $B$ and $S$.

The triangle $\Delta ABT$ has an angle with measure $180^\circ-12^\circ$ in $A$, and two $6^\circ$ angles in $B$ and $T$. Then $D$ is determined by constructing the parallelogram $TACD$.

Note that the circle $(A)$ passes through $O$.

We draw $DA$ and ask for the angle between the lines $AD$ and $BCD$. The picture suggests the next step, try to pause consuming the solution at this point...


(2) Some more observations and another picture:

In the regular polygon $P_{30}$ recall that the vertices are indexed via $\Bbb Z/30$. Then $BA=05$ or in general a segment $[k,k+5]$ is the side of a regular hexagon inscribed in $(O)$. So $BA$ is also equal to the radius $OA=OB$ in $(O)$. The OP mentions the need to find an equilateral triangle, well, here is one...

We consider closer the relation of the circles centered in $O,A,C$ having as radius this distance $$ OA=OB=AB=AS=AT=CD=CO\ . $$ Consider also the regular congruent $30$-gons on the circles $(O)$, $(A)$, $(C)$. (Note that main diagonals in the one or the other $30$-gon pass through vertices of the other $30$-gons, angle argument.)

Observation: We obtain the $30$-gon inside $(C)$ from the $30$-gon inside $(A)$ by a translation with the vector $\vec{AC}=\vec{TD}$.

mathematics stackexchange picture for problem 3911812, fig2

This picture is of course too much and not needed, for the OP we want to know the angles in the paralellogram $ACDT$ formed by the diagonal $DA$ with the sides. Let $U$ be the point with $\vec{AC}=\vec{SU}$. The parallelograms $TDCA$ and $ACUS$ are congruent.


(3) An other observation and yet an other picture:

In (2) the way to reason was as follows. The circles $(A)$ and $(C)$ have the same radius, thus they are congruent. Consider the vector $\vec{AC}$, it translates $(A)$ into $(C)$, and in particular also a specific $30$-gon inscribed in $A$ into a specific $30$-gon inscribed in $(C)$ if we insure that (at least) one vertex of the first $30$-gon goes to one vertex of the other $30$-gon.

The same applies also with the pair of circles $(A)$ and $(O)$, and/or with the pair of circles $(C)$ and $(O)$ correspondingly instead.

A similar observation can be done with rotations (instead of translations). For example... The circles $(A)$ and $C$ are passing through $O$ (and an other point). Consider the rotation centered in $O$ which maps $A$ into $C$. Then it maps the $30$-gon inscribed in $A$ with one vertex in $O$ into the $30$-gon inscribed in $C$ with one vertex in $O$. A picture suggesting the movement of the vertices by drawn circles of guidance would be:

mathematics stackexchange picture for problem 3911812, fig3

Again, this is too much for our purposes, but we are only collecting ideas.


When searching for a solution, the idea is to use some pointed points under the many points constructed above.


(4) Solution I:

  • Let $O$ be a point and $(O)$ a circle centered in $O$ with radius $R$. Let $O(30)$ denote a fixed $30$-gon inscribed in $(O)$, its vertices are denoted by $0,1,2,\dots\in\Bbb Z/30$, and for some points we use alternative notations.
  • The first three points we want to evidentiate are $B=0$, $C=3$, and $A=5$.
  • Let $(A)$ and $(C)$ be the circles centered in $A$, respectively $C$ passing through $O$. They also have radius $R$. We draw the $30$-gons $A(30)$, and respectively $C(30)$ inscribed in these two circles having $O$ as one vertex in common, and denote the vertices in them by $A_0$, $A_1$, $A_2$, and so on, respectively $C_0$, $C_1$, $C_2$, and so on. The indices are also living in $\Bbb /30$.
  • The starting point when counting is $O$, so $(A)$ and $(C)$ intersect in $O=A_0=C_0$, and in $A_{13}=C_{-13}$.
  • $\Delta ABO$ is equilateral, its sides are $R$.
  • We thus know the angle $\widehat{OBC}=60^\circ+12^\circ$, so $BC$ intersects $(C)$ for the second time in $D=C_{-9}$.
  • The radius $AB$ sits on the diameter $A_5A_{-10}$. We know the angle $\widehat{ABC}=12^\circ$, so $BC$ intersects for the second time $(A)$ in $A_{-12}$.
  • Similar knowledge of angles shows that the parallel through $A$ to the line $BCDA_{-12}$ is the line $TC_{-8}AA_4C_5$.

mathematics stackexchange picture for problem 3911812, fig4

  • Now start the solution, and show the equality of lines $(5,-5)$ and $C_3C_{-9}$. In $(O)$, the line between $5$ and $-5$ is the perpendicular bisector of the radius $O0=OB$, as a visual help consider the equilateral triangles $\Delta OB5=\Delta OBA$ and $\Delta OB-5$. The translation of the plane with vector $\vec{OC}$ transforms circles $(O)\to(C)$, and regular polygons $O(30)\to C(30)$. The chord $(5, -5)$ is mapped to the chord $C_{-13}C_7$. Note also that the perpendicular in $0=A_5$ to $(O)$ has known angle w.r.t. $AB=AA_5$, wo we can identify the direction perpendicular to $OB$ also in the circle $(A)$, it is $A_5A_{15}$. From this we obtain a parallelism $$(5,-5)\| C_{-13}C_7\|C_{-13+4}C_{7-4}=C_{-9}C_3\|\dots \\ \dots\| A_5A_{-15}\|A_4A_{-14}\|A_3A_{-13}\|A_2A_{-12}\|A_1A_{-11}\|OA_{-10}\ .$$ The diameter through $C_{-9}=D$ in $C$ is $C_{-9}C_6$. Reflecting the line $$ (l) =C_{-9}CC_6=DCB=\color{blue}{A_{-12}A_5} $$ w.r.t. to the line $\color{red}{C_{-9}C_3}$ gives the line $$ (l')=C_{-9}C_{3-3}=DO=D\color{blue}{A_0}\ . $$ And yes, there is exactly one isosceles trapez $\color{blue}{A_{-12}A_0A_5A_?}$ which is symmetric w.r.t. a direction matching the above parallelism, it is $\color{blue}{A_{-12}A_0A_5A_{-13}}$, and its symmetry axis passes through the center $A$ of the circle $(A)$. So the parallel lines $\color{red}{C_{-9}C_3}$ and $\color{red}{-5\;A}$ both contain $A$, so they are equal. This shows the colinearity of the points $$ D=C_{-9}\ ,\ A=5\ ,\ C_3\ ,\ -5\ . $$ The needed angle is for instance half of the angle between the diagonals of the above trapez, so it is $$ \widehat{ADB} =\frac 12\widehat{ODB} =\frac14(\overset\frown{A_{-12}A_{-13}} + \overset\frown{A_0A_5}) =\frac14(1\cdot 12^\circ + 5\cdot 12^\circ ) =18^\circ\ . $$

$\square$


(5) Solution II:

We use the same notations as above, else this would make 60% of the solution.

Using the rotation $\mathcal R$ centered in $O$ that maps $C=3$ to $A=5$ we can ask for the angle in $A_{-9}$ in $\Delta A_{-9}57=\mathcal R\Delta C_{-9}35$ instead of the angle in $A_{-9}$ in $\Delta C_{-9}35=\Delta DCA$ asked in the OP, and thus we completely remove from the picture $(C)$ and $C(30)$.

mathematics stackexchange picture for problem 3911812, fig5

We know the angle $\widehat{A_{-9}AO} =\widehat{C_{-9}CO}$, or use the colinearity $(A_{-9}52)=\mathcal R(C_{-9}30)$, to obtain that $A_{-9}52A_6$ is a line (supported on the diameter $A_{-9}A_6$.

The angles between the one fixed line $O,10$ and $OA_{-9}$ and $O8$ are equal, so $O, 8,A_{-9}$ are colinear. The two angles

  • between the chords $8,O,-7$ and $7,-3$ in $(O)$
  • between $A_{-9}A_0$ and $A_{-9}A_3$ in $(A)$

are equal, $4-1=3$, and two of the above chords are on the same line, so we obtain the colinearity of points $$ A_{-9},\ A_3\ ,\ 7\ ,\ -3\ . $$ The needed angle is thus $$ \widehat{CDA} = \widehat{3C_{-9}5} \overset{\mathcal R}= \widehat{5A_{-9}7} = \widehat{A_6A_{-9}A_3} = \frac 12\overset\frown{A_6A_3} =\frac 12\cdot3\cdot 12^\circ =18^\circ \ . $$ $\square$

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  • $\begingroup$ Thank you, I was looking for something like this. $\endgroup$ – Tan Nov 20 '20 at 9:11
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While I was trying to correctly plot the angles in $\triangle ABC$ I figured out that an easy way is to use a regular triangle and a regular pentagon.

In the figure below you can see that, drawing the regular triangle $\triangle BCE$ on $BC$ and a regular pentagon on $BE$, the vertex of the pentagon is indeed in $A$. In fact it is known that $\angle EAB=36°$ and $\angle ABE=72°$ (you can easily verify it using symmetries in the pentagon), so $\angle ABC=\angle ABE-\angle CBE=72°-60°=12°$ and $\angle CAB=18°$ as it bisects $\angle EAB$.

solution

When I see pentagons, what comes to my mind are golden triangles and golden ratios: $\triangle ABE$ is a golden triangle so $AB$ and $BC=BE$ are in golden ratio.

So we can build the golden triangle $\triangle KBC$ over $BC$ with $BK=CK=AB$. But the nice thing about golden triangles is that they "sum up": the triangle with $BD=BC+CD$ as a side and $BK$ as base is again a golden triangle as $\angle KBD=\angle KBC=72°$ and $BD$ and $BK$ are in golden ratio. So $DK=BD$ and $\angle BDK=36°$.

Now we can use again an equilateral triangle: $AB=BK$ and $\angle ABK=60°$ so $AK=AB$, the triangles $\triangle ABD$ and $\triangle ADK$ are equivalent as their sides have the same length, and finally $\angle BDA$ bisects $\angle BDK$.

So the solution is $\angle BDA=18°$.

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Use law of sines. In $\triangle ABC$: $$\frac{\sin(180^\circ-12^\circ-18^\circ)}{AB}=\frac{\sin 12^\circ}{AC}$$ In $\triangle ACD$: $$\frac{\sin(180^\circ-12^\circ-18^\circ-x)}{AB}=\frac{\sin x}{AC}$$ In the last equation I've used $CD=AB$. Therefore, take $AB/AC$ from both equations: $$\frac{\sin 150^\circ}{\sin 12^\circ}=\frac{\sin (150^\circ-x)}{\sin x}$$ Use $$\sin(150^\circ-x)=\sin 150^\circ\cos x-\cos 150^\circ\sin x$$ to get a simple equation for $\tan x$.

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  • $\begingroup$ Sorry, I'll fix the typo. Thanks @CalvinLin $\endgroup$ – Andrei Nov 17 '20 at 21:54
  • $\begingroup$ Thank you for your answer, can you think of a solution which uses "geometrical" methods like drawing line segments or forming new triangles etc.? I forgot to add that to the question but such an answer would be great. $\endgroup$ – Tan Nov 17 '20 at 22:01

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