Solve for the angle x?

Question

The problem seems really impossible to me, despite solving many problems on triangles. So I hope that you can solve it.

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Answer 1

Unfortunately, some case analyses and contradiction things must be done and I wasn't able to come up with any nice and clean way.

First create triangle $ACG$ congruent to $ABC$ and label $F$ as shown in the picture. Mirror $D$ along axis $AC$ to point $P$. Connect necessary points as shown in the picture.

In trapezoid $ADCG$, triangles $ACG$ and $DGC$ are symmetric so congruent as well. So $DGC$ is also a triangle congruent to $ABC$ where angle $\angle DGC=x$

At the same time, $PA = AD = EC, AG = CD, \angle PAG = 90 - {x\over 2} - x = \angle ECD \implies \triangle PAG \cong \triangle ECD$.

Furthermore it's easy to see $DE=PG$ and $\triangle DEG \cong \triangle GPC$.

Now let $\angle GDE = \angle CGP = a$.

With some angle tracing (with the known congruences every single angle can be expressed in terms of x), $a = 40 - {x\over 2}$, and $\angle DEP = 280 - 3x$.

Case $1$: $EP$ is not parallel to $DG$ and intersect $DG$ on the extension ray from $D$ to $G$. Then this means $\angle GDE + \angle DEP < 180$ which means $320 - {7x\over 2} < 180$ so $x > 40$ and $a < 20$.

Therefore $\angle DGP = x - a > 20 > a = \angle GDE$. However since $DE = PG$ and $\angle DGP > \angle GDE$, $P$ would have a further distance to $DG$ and $E$ and $PE$ would intersect $GD$ on the extension ray from $G$ to $D$. contradiction.

Case $2$: $EP$ is not parallel to $DG$ and intersect $GD$ on the extension ray from $G$ to $D$. Then this means $\angle GDE + \angle DEP > 180$ which means $320 - {7x\over 2} > 180$ so $x < 40$ and $a > 20$. All similar thing and we can conclude $\angle GDE > \angle DGP$ so $EP$ must intersect $DG$ on the extension from $D$ to $G$. contradiction.

So $EP$ must be parallel to $DG$ and $320 - {7x\over 2} = 180$ and $x=40$.

Answer 2

With the notations on the figure :

$X=\widehat{BAC}$

$\beta=\widehat{CBA}=\frac{\pi-X}{2}$

$\gamma=\widehat{DCB}=\pi-2\beta=X$

$\delta=\widehat{ACD}=\beta-\gamma=\frac{\pi-3X}{2}$

Without lost of generality, let $BC=1$

$BD=AE=2\cos(\beta)=2\sin(\frac{X}{2})$

$BA=\frac{1}{2\cos(\beta)}=\frac{1}{2\sin(\frac{X}{2})}$

$CE=DA=BA-BD=\frac{1}{2\sin(\frac{X}{2})}-2\sin(\frac{X}{2})$

$EH=CE\sin(\delta)= (\frac{1}{2\sin(\frac{X}{2})}-2\sin(\frac{X}{2})) \cos(\frac{3X}{2})$

$CH=CE\cos(\delta)= (\frac{1}{2\sin(\frac{X}{2})}-2\sin(\frac{X}{2}))\sin(\frac{3X}{2})$

$HD=CD-CH=1-\frac{\sin(\frac{3X}{2} )}{2\sin(\frac{X}{2})}+2\sin(\frac{X}{2})\sin(\frac{3X}{2})$

$\tan(Y)=\tan(\widehat{CDE})=\frac{EH}{HD}=\frac{(\frac{1}{2\sin(\frac{X}{2})}-2\sin(\frac{X}{2})) \cos(\frac{3X}{2})}{ 1-\frac{\sin(\frac{3X}{2} )}{2\sin(\frac{X}{2})}+2\sin(\frac{X}{2})\sin(\frac{3X}{2}) }$

The relationship between angles $X$ and $Y$ is : $$\tan(Y)=\frac{(1-4\sin^2(\frac{X}{2})) \cos(\frac{3X}{2})}{2\sin(\frac{X}{2}) -\sin(\frac{3X}{2} )+4\sin^2(\frac{X}{2})\sin(\frac{3X}{2}) } $$

$$\cot(Y)=\frac{2\sin(\frac{X}{2}) } {(1-4\sin^2(\frac{X}{2})) \cos(\frac{3X}{2})} -\tan(\frac{3X}{2}) \tag 1$$

The function $Y(X)$ is represented on the next figure.

For example, if we set $X=\pi\frac{40}{180}$ that is an angle of $40°$ , the numerical result is : $$Y\simeq 0.872664625997165\quad\to\quad 180\frac{Y}{\pi}\simeq 50.00000000000003$$

Thus, from numerical calculus, at $Y=50°$ corresponds $X=40°$ with a very good accuracy.

Analytically one have to prove that the equation $(1)$ returns exactly $Y=\pi\frac{50}{180}$ for $X=\pi\frac{40}{180}$ .

Consider la function : $$f(X,Y)=\frac{2\sin(\frac{X}{2}) } {(1-4\sin^2(\frac{X}{2})) \cos(\frac{3X}{2})} -\tan(\frac{3X}{2}) -\cot(Y) $$

we have to prove that this function is $=0$ when $Y=\pi\frac{50}{180}$ and $X=\pi\frac{40}{180}$

$$f(\frac{2\pi}{9},\frac{5\pi}{18}) =\frac{2\sin(\frac{\pi}{9}) } {(1-4\sin^2(\frac{\pi}{9})) \cos(\frac{\pi}{3})} -\tan(\frac{\pi}{3}) -\cot(\frac{5\pi}{18}) $$

$$f((\frac{2\pi}{9},\frac{5\pi}{18}) =\frac{4\sin(\frac{\pi}{9}) } {(1-4\sin^2(\frac{\pi}{9}))} -\sqrt{3} -\cot(\frac{5\pi}{18}) $$

Let $s=\sin(\frac{\pi}{18})$

$\sin(\frac{\pi}{9})=2s\sqrt{1-s^2}$

$\cos(\frac{5\pi}{18})= (16s^4-12s^2+1)\sqrt{1-s^2}$

$\sin(\frac{5\pi}{18})=(16s^5-20s^3+5s)$

$$f(\frac{2\pi}{9},\frac{5\pi}{18}) =\frac{8s\sqrt{1-s^2} } {1-4(2s\sqrt{1-s^2})^2} -\sqrt{3} -\frac {(16s^4-12s^2+1)\sqrt{1-s^2}}{16s^5-20s^3+5s} $$

In fact, we have to check if one of the roots of the equation : $$\frac{8s\sqrt{1-s^2} } {1-4(2s\sqrt{1-s^2})^2} -\sqrt{3} -\frac {(16s^4-12s^2+1)\sqrt{1-s^2}}{16s^5-20s^3+5s}=0 \tag 2$$ is equal to $\sin(\frac{\pi}{18})$ .

Expanding and factoring transform the equation to : $$(8s^3-6s-1)(8s^3-6s+1)(4096s^{12}-13312s^{10}+16128s^8-8832s^6+2096s^4-176s^2+1)=0$$

For $s=\sin(\frac{\pi}{18})$ the polynomials $8s^3-6s+1\neq 0$ and $4096s^{12}-13312s^{10}+16128s^8-8832s^6+2096s^4-176s^2+1\neq 0$ .

Only the equation : $$8s^3-6s+1=0$$ is likely to have a root around to $\sin(\frac{\pi}{18})$. Then we have to prove that it is the case.

We put into the equation $s=\sin(\frac{\pi}{18})=\frac{\exp(i\frac{\pi}{18})-\exp(-i\frac{\pi}{18})}{2i}$ $$8s^3-6s+1=8\left(\frac{\exp(i\frac{\pi}{18})-\exp(-i\frac{\pi}{18})}{2i} \right)^3-6\left(\frac{\exp(i\frac{\pi}{18})-\exp(-i\frac{\pi}{18})}{2i} \right)+1$$ After expansion and simplification all terms cancel. We get : $8s^3-6s+1=0$ .

Thus $s=\sin(\frac{\pi}{18})$ is a root of equation $(2)$ and as a consequence $f(\frac{2\pi}{9},\frac{5\pi}{18}) =0$ ,

This is the analytical proof that $X=40°$ and $Y=50°$ is an exact solution.

Answer 3

Here is a synthetic solution.

Let $\angle BAC = x$ and $\angle DCE = y$ Now, $\angle ABC = 60 + \frac{y}{3}$. If we prove $y= 30$, the proof would be complete. Let $BC = a$ Note that $$BD = 2a \cos ({60 +\frac{y}{3}})$$ and since $AB= \frac {a}{2\cos ({60 + \frac{y}{3}})}$, we can calculate after simplifying and using the property :$\frac {4\cos^2 t -1}{2\cos t} = \frac {3- 4\sin^2 t}{2\cos t}= \frac {\sin 3t}{\sin 2t}$ $$\frac {AD}{a} = \frac {\sin y}{\cos ({30 + \frac {2y}{3}})}$$ By sine rule in $\Delta DEC$, and using the above result,

$$\sin{y}\sin({y+50})= \sin{50}\cos({30+\frac{2y}{3}})$$

And so,

$$2\sin{y}\sin({y+50})= 2\sin{50}\cos({30+\frac{2y}{3}})$$

if $y=30$, then this equation is satisfied because RHS becomes $\sin 100$ and LHS becomes $\sin 80$ and they are obviously equal. Therefore, $x=40$

Note: To see that that equation has no other solutions than $y=30$, I suggest using Wolfram Alpha.