Adventitious Quadrangle 2

This solution will be attempted without trigonometry.

First, some simple angle chasing. Also its not hard to see there is an isosceles triangle. Now, it seems like we are stucked. But If we look long enough, the 80°, 100° supplementary angles are looking more and more suspicious.

The first trick: what if we "move" two of the triangles such that two of the equal sides coincides? A cyclic quadrilateral is formed. (Why cyclic?)

Next, draw another diagonal and hunt down all angles. (Inscribed angle theorem.) But what is the use of this cyclic quadrilateral?

The second trick is to realise there is a pair of congruent triangle hidden. The purple triangles are congruent. (Why so?)

$\therefore x = 40^\circ$

First, fill in the angles that can be determined by triangle sum and straight angles, and also label the vertices $A$ , $B$ , $C$ , $D$ , and $E$ as follows:

Since $\angle EAD = \angle EDA = 30°$ , $\triangle EAD$ is an isosceles triangle, and $EA \cong ED$ .

Next, construct equilateral triangle $\triangle FED$ , with $C$ on $EF$ , and fill in the angles:

Since $FD \cong ED$ by equilateral triangle sides, and $ED \cong EA$ (from above), $FD \cong EA$ . Since $\angle BEA = \angle CFD = 60°$ and $\angle BAE = \angle CDF = 40°$ , $\triangle DFC \cong \triangle AEB$ by ASA congruence. Therefore, $BE \cong CF$ .

Finally, construct equilateral triangle $\triangle ECG$ , with $G$ on $ED$ , and fill in the angles:

Since $EC \cong EG$ by equilateral triangle sides, and $EF \cong ED$ by equilateral triangle sides, $EF - EC \cong ED - EG$ , or $CF \cong GD$ . From above, $BE \cong CF$ , so $BE \cong GD$ . Since $CE \cong CG$ by equilateral triangle sides, and $\angle CEB = \angle CGD = 120°$ , $\triangle BEC \cong \triangle DGC$ by SAS congruence. Therefore, $\angle BCE = \angle DCG = 40°$ , which means $x = \boxed{40}°$ .

Solution using trigonometry.

Consider isosceles $\triangle ADE$ . Let $AE=ED=1$ . Apply cosine law.

$(AD)^2=(AE)^2+(ED)^2-2(AE)(ED)(\cos 120)$

$(AD)^2=1+1-2(\cos 120)$

$AD \approx 1.73$

Consider $\triangle ABD$ . Apply sine law.

$\dfrac{AB}{\sin 30}=\dfrac{AD}{\sin 80}$

$AB=\dfrac {\sin 30}{\sin 80}(1.73) \approx 0.88$

Consider $\triangle ACD$ . Apply sine law.

$\dfrac{AC}{\sin 50}=\dfrac{AD}{\sin 100}$

$AC = \dfrac{\sin 50}{\sin 100} (1.73) \approx 1.35$

Consider $\triangle ABC$ . Apply cosine law.

$(BC)^2=(0.88)^2+(1.35)^2-2(0.88)(1.35)(\cos 40)$

$BC \approx 0.88$

Since $BC = AB$ , $\triangle ABC$ is isosceles. Therefore,

$x=\angle BAC = \boxed{40^\circ}$

Some solutions really opened up my vision towards Geometry. I will give a rather mundane approach, which I exploit in most of the angle-chasing problems.

@David Vreken forgive me for using your image. I totally forgot how to draw these neat images online.

Considering $\triangle ABC$ , we have from sine-formula, $\dfrac{\sin x}{AB} = \dfrac{\sin 40}{BC}$ .

Again, in $\triangle DBC$ , we have $\dfrac{\sin 20}{BC} = \dfrac{\sin (100 + x)}{BD}$ .

Finally, in $\triangle ABD$ , we have $\dfrac{\sin 70}{BD} = \dfrac{\sin 30}{AB}$ .

We see that, on multiplying the above relations likewise, the denominators cancel out. That is the motivation to use this approach.

So we get,

$\sin x \cdot \sin20 \cdot \sin70 = \sin 40 \cdot \sin(100+x) \cdot \sin30$

$\Rightarrow \sin x \cdot \sin20 \cdot \sin70 = (2 \cdot \sin20 \cdot \cos20) \cdot \sin(100+x) \cdot \dfrac{1}{2}$

Remembering, $\sin70 = \cos20$ and cancelling common terms on both sides, we are left with

$\sin x = \sin(100+x)$

$\Rightarrow \sin x - \sin(100+x) = 0$

$\Rightarrow 2 \cdot \cos(x+50) \cdot \sin(-50) = 0$

$\Rightarrow \cos(x+50) = 0$ $($ Since $\sin50^\circ \neq 0$ $)$

$\Rightarrow x = \boxed{40^\circ} .$

By creating a symmetrical triangle based on ADC, reflected along AD, angle x can be ascertained from resulting triangle ABC'.

Let $E=AC \cap BD$ and $P=AD\cap BC$ . $DPCE$ is a cyclic quadrilateral, then $\angle DCE=\angle DPE$ . By Ceva's Theorem $\frac{\sin \angle BAE}{\sin \angle EAP}\cdot\frac{\sin \angle APE}{\sin \angle EPB}\cdot\frac{\sin \angle PBE}{\sin\angle EBA}=1$ $\frac{\sin 30^{\circ}}{\sin 40^{\circ}}\cdot\frac{\sin \angle APE}{\sin EPB}\cdot\frac{\sin 20^{\circ}}{\sin 30^{\circ}}=1$ Finally $\angle APE+\angle EPB=60^{\circ}$ and $\frac{\sin \angle APE}{\sin 40^{\circ}}=\frac{\sin\angle EPB}{\sin 20^{\circ}}$ implies $\angle APE=40^{\circ}$ .

This will be another way to construct auxiliary lines. After looking long enough at the diagram, it feels very tempting to show somehow that A=F or B=F, and proving either of these will give the value of the angle.

So the natural goal is to produce as many isosceles or equilateral triangles as possible, using side lengths the length of either A, B, or F. Here's how I did:

First construct C. Now the triangles between C-D and E-B are identical, so C=B. The 80° angles give A=C, therefore A=B. Next, move B closer to A, creating a parallelogram BFB'F' and an equilateral triangle AB'G. $(\angle AB'=\angle AB=180^{\circ}-70^{\circ}-50^{\circ}=60^{\circ})$

By the inscribed angle theorem, B', F', and G form a set of radii of a same circle, thus B'=F'.

Immediately A=B=F, where the pair A=F can give the desired $\boxed{x=40°}$ .

Geometry solution:

Draw equilateral triangle ABE . See sketch below for the positions of points D,F,C,G,E and values of angles ∠ (DAE)= 10, ∠(CBE)=10, ∠(ACB)=100,∠(DCE) =60,∠ (ADB)=80 Since AC and BD are angle bisector of ∠(BAE) and ∠(ABE) so AC is the perpendicular bisector of BẸ Triangle ACE congruent to triangle ACB ( case SAS) so ∠(ACB)= ∠(ACE)= 100 degrees and x= 100-60= 40 degrees

Answer using trig:

Firstly, I used the properties of an isosceles triangle to take both AE and BE as 1 unit in length. I then used basic angle rules to obtain all angles labelled.

Using the sine rule twice to find some useful side lengths:

$\frac{EC}{sin(20)}=\frac{1}{sin(100)}$

$EC \approx 0.347$

$\frac{DE}{sin(60)}=\frac{1}{sin(80)}$

$DE \approx 0.653$

Using the cosine rule to find the final side length:

$(CD)^{2}=(ED)^2+(EC)^2-(2\times ED \times EC \times \cos 120)$

$CD \approx 0.879$

Then used the sine rule once more to find angle x:

$\frac{sin (x)}{0.653}=\frac{sin (120)}{0.879}$

$\therefore x = 40^{\circ}$

Like everyone else, I started with angle hunting.

First note that $120 + x + \alpha = 180$ and thus, $x + \alpha = 60$

Using the sine law in triangles DMA and CMB,

$AM = DM \frac{\sin (80)}{\sin (40)}$

$BM = CM \frac{\sin(100)}{\sin(20)}$

Since triangle AMB is isoceles, $AM = BM$

$DM \frac{\sin (80)}{\sin (40)} = CM \frac{\sin(100)}{\sin(20)}$

$\frac{DM}{CM} = \frac{\sin(100) \sin(40)}{\sin(20) \sin(80)}$

80 and 100 are complementary angles, so $\sin 80 = \sin 100$ and they cancel out in the fraction.

Resulting: $\frac{DM}{CM} = \frac{\sin(40)}{\sin(20)}$

Using the sine law again but in triangle CMD,

$\frac{DM}{\sin x} = \frac{CM}{\sin \alpha}$

$\frac{DM}{CM} = \frac{\sin x}{\sin \alpha}$

$\frac{\sin x}{\sin \alpha} = \frac{\sin(40)}{\sin(20)}$

At this point it looks like we're stuck. It's going to be impossible to guess a solution from this.

However, the solution has been right there before our eyes the whole time. Remember $x + \alpha = 60$ ? Well, apparently $40 + 20 = 60$

And that also a solution to $\frac{\sin x}{\sin \alpha} = \frac{\sin(40)}{\sin(20)}$ , so...

Therefore, $x = 40^{\circ}$

Use a 18-gon for this one. Start by drawing lines $RT$ , $PA$ and $QC$ , so that triangle $\Delta RPC$ is an equilateral triangle, and therefore triangle $\Delta QRC$ is isosceles, with base angles $40$ degrees. Next, we draw lines $CB$ , $PB$ and $AB$ , so that angle $\angle APB = 80$ degrees, and angle $\angle ABP = 70$ degrees. From this we determine that angle $\angle CBP = 20$ degrees, and finally that angle $\angle PCB = 80$ degrees, so that not only triangle $\Delta BCP$ is isosceles, line segments $QC$ and $CB$ are colinear.

Therefore $AD = PT = CT = DC$ . From this, we see that triangle $\Delta ADC$ is isosceles, and that angle $\angle DCA = 40$ .

This proof requires showing that lines $RT, PA, QB, SD$ intersect at the same point $C$ . See the following:

Here, we'll only look at the question of the concurrency of lines $AD$ , $BE$ , $CF$ , all meeting (or not) at point $Q$ . Those lines, as well as line $AB$ , are drawn in the 18-gon as follows:

First we draw the symmetrical kite $ABCQ$ , formed by reflection of triangle $\Delta ABC$ which has angles $10, 140, 30$ degrees. Thus, we know that not only both triangles $\Delta ABQ$ , $\Delta BCQ$ are isosceles, $\Delta BCQ$ is equilateral.

Then if lines $BQ$ and $BE$ both make the same angle $60$ from line $BC$ , then both lines are colinear.

Likewise for lines $CQ$ and $CF$ for the same reason.

Finally, if lines $AQ$ and $AD$ both make the same angle $20$ from line $AB$ , then they are colinear as well.

This establishes the concurency of lines $BE$ , $CF$ , and $AD$ , meeting at point $Q$ .

I feel writing this solution as this one is really incredible!!! credits : Dong Kwan Choo

We have, $\frac{AB}{AD}\frac{BC}{AB}\frac{CD}{CB}\frac{DA}{CD}=1$ so by sine rule, $\frac{\sin40}{\sin x}\frac{\sin(60-x)}{\sin20}\frac{\sin100}{\sin30}\frac{\sin30}{\sin80}=1$ $or$ $\frac{\sin(60-x)}{\sin x}=\frac{1}{2} \sec20$ $or$ , $\cot x=\frac{\sqrt{3}}{1+\sec20}$ from above equation sove for $x$ to get $x=40$

Draw an accurate diagram.

I used brute force. The lower left corner is (0,0) and the lower right corner is (1,0). The upper left corner is (x1,y1) and the upper right corner is (x2,y2). The equations are y1/x1=tan 70° , y1/(1-x1)=tan 30° ; y2/x2=tan 30° ; y2/(1-x2)=tan 50° ; The slope of the upper line is (y2-y1)/(x2-x1), The angle is -10°. So x=40°. The J-code is this.

't3 t5 t7'=.3 o. o. 180%~ 30 50 70

y1=.t7*x1=.%1+t7%t3

y2=.t3*x2=.%1+t3%t5

(30+30)-30+t^:_1 (y2-y1)%x2-x1

40

It's about thinking outside the box: $ABC$ is an equilateral triangle (the side length doesn't matter since everything can be scaled to 1) with bisectors $CD$ and $AE$ . Now consider the affine transformation that rotates 10° clockwise and scales to $BG:BE$ . This maps $E$ to $G$ and $D$ to $F$ . Thus the segment $FE$ forms a $10°$ angle to $DE$ . Now our original configuration is simply $AFGC$ and the angle $x$ is just the (clockwise oriented) angle formed by $EA$ to $GF$ , thus we get

$x=\widehat{(EA,GF)}=\widehat{(EA,ED)}+\widehat{(ED,GF)}=30°+10°$

$x=40°$

I'm guessing there are a lot more nice relationships that follow from this type of strategy

I found what could be coincidental, but if not is just an extremely simple solution.

Lets call the centre O, the point of angle 50° we shall call A, and the point of angle 70° will be called B (therefore for example (here (<) denotes an angle $\angle OAB = 30 ^ \circ = \angle OBA$ ).We will also call angle x <CDO (top left=C, top right=D, centre=O, bottom left=B and bottom right=A.

So, with simple geometry rules w find that: <OAB = 30 = <OAB (making OAB an isosceles triangle)

<BOA = 120 = <COD (Vertically opposite angles are equal (VO))

<COB = 60 = <DOA (((360 -(120*2))/2 = 60) and therefore VO)

<BCO = 80 ( Angles in a two dimensional triangle add to 180° (AT))

<ODA = 100 (AT)

Therefore we get that 180 - 120 = <OCD + <ODC, (or x° + y° = 60°)

We can also get that 360 - ( <ODA + <DAB + <ABC + <BCO (100+50+70+80 = 300) = 60.

This is where I made my (likely coincidentally correct) assumption. Based on the fact that ABCD would be a regular trapezium (internal angle = 360), it is almost as if point A has been stretched until <CBO is double <OAD (we can ignore the triangle OBA as they (<CBA and <DAB) share the common angle of 30°). From this I said that because <CBO = 2*<OAD, then x = 2y (I don't quite know how to explain my thought process but it seemed to make sense) so, as x+y = 60, then 3y=60 and y=20, therefore 2(20) = x = 40.

Hence \I found what could be coincidental, but if not is just an extremely simple solution.

Lets call the centre O, the point of angle 50° we shall call A, and the point of angle 70° will be called B (therefore for example (here (<) denotes an angle <OAB = 30° = <OBA).We will also call angle x <CDO (top left=C, top right=D, centre=O, bottom left=B and bottom right=A.

So, with simple geometry rules w find that: <OAB = 30 = <OAB (making OAB an isosceles triangle)

<BOA = 120 = <COD (Vertically opposite angles are equal (VO))

<COB = 60 = <DOA (((360 -(120*2))/2 = 60) and therefore VO)

<BCO = 80 ( Angles in a two dimensional triangle add to 180° (AT))

<ODA = 100 (AT)

Therefore we get that 180 - 120 = <OCD + <ODC, (or x° + y° = 60°)

We can also get that 360 - ( <ODA + <DAB + <ABC + <BCO (100+50+70+80 = 300) = 60.

This is where I made my (likely coincidentally correct) assumption. Based on the fact that ABCD would be a regular trapezium (internal angle = 360), it is almost as if point A has been stretched until <CBO is double <OAD (we can ignore the triangle OBA as they (<CBA and <DAB) share the common angle of 30°). From this I said that because <CBO = 2*<OAD, then x = 2y (I don't quite know how to explain my thought process but it seemed to make sense) so, as x+y = 60, then 3y=60 and y=20, therefore 2(20) = x = 40.

Hence x = 40°