Any point $D$, which is interior to the triangle $\triangle ABC$, determines vertex angles having the following measures:
$m(BAD) = x,$
$m(ABD) = 2x,$
$m(BCD) = 3x,$
$m(ACD)= 4x,$
$m(DBC) = 5x.\\$
Find the measure of $x$. (Answer:$10^\circ$)
My progress:
$\dfrac{AB}{AD}=\dfrac{\sin(180−3x)}{\sin(2x)}\\ \dfrac{AC}{AD}=\dfrac{\sin(11x)}{\sin(4x)} AB=AC \implies \dfrac{\sin(3x)}{\sin(2x)}=\dfrac{\sin(11x)}{\sin(4x)}$
but it is a complex equation to solve. Is there another way?
On
Recall that the interior angles of any triangle always add up to $\pi$. From this, we can determine:
$$\angle CAD = \pi - 15x$$ $$\angle ADB = \pi - 3x$$ $$\angle BDC = \pi - 8x$$ $$\angle ADC = 11x$$
Applying the Law of Sines to each of the triangles in the figure gives:
$$\triangle ABC \implies \frac{AB}{\sin(7x)} = \frac{AC}{\sin(7x)} = \frac{BC}{\sin(\pi - 14x)}$$ $$\triangle ABD \implies \frac{AB}{\sin(\pi - 3x)} = \frac{AD}{\sin(2x)} = \frac{BD}{\sin(x)}$$ $$\triangle ACD \implies \frac{AC}{\sin(11x)} = \frac{AD}{\sin(4x)} = \frac{CD}{\sin(\pi - 15x)}$$ $$\triangle BCD \implies \frac{BC}{\sin(\pi - 8x)} = \frac{BD}{\sin(3x)} = \frac{CD}{\sin(5x)}$$
Note that because all angles must be positive, we must have $0 < x < \frac{\pi}{15} = 12°$.
These can be slightly simplified by using the identity
$$\sin(\pi - \theta) = \sin(\pi)\cos(\theta) -\cos(\pi)\sin(\theta) = 0\cos(\theta) - (-1) \sin(\theta) = \sin(\theta)$$
WLOG, assign the equal sides ($AB$ and $AC$) of $\triangle ABC$ a length of “1”. Then, the above Law of Sines equations let us find the lengths of the other sides. Most have two different expressions for them.
$$AD = \frac{\sin(2x)}{\sin(3x)} = \frac{\sin(4x)}{\sin(11x)}$$ $$BC = \frac{\sin(14x)}{\sin(7x)}$$ $$BD = \frac{\sin(x)}{\sin(3x)} = \frac{\sin(3x)\sin(14x)}{\sin(8x)\sin(7x)}$$ $$CD = \frac{\sin(15x)}{\sin(11x)} = \frac{\sin(5x)\sin(14x)}{\sin(8x)\sin(7x)}$$
You've already found the equation based on line $AD$, but let's another way and base it on $BD$. We can simplify it somewhat by using the identity $\sin(2\theta) = 2\cos(\theta)\sin(\theta)$.
$$\frac{\sin(x)}{\sin(3x)} = \frac{2\sin(3x)\cos(7x)}{\sin(8x)}$$ $$\sin(x)\sin(8x) = 2\sin^2(3x)\cos(7x)$$
For brevity, let $u = e^{ix}$. Then $\sin(nx) = \frac{u^n - u^{-n}}{2i}$ and $\cos(nx) = \frac{u^n + u^{-n}}{2}$.
$$(\frac{u - u^{-1}}{2i})(\frac{u^8 - u^{-8}}{2i}) = 2(\frac{u^3 - u^{-3}}{2i})^2(\frac{u^7 + u^{-7}}{2})$$ $$\frac{-1}{4}(u - u^{-1})(u^8 - u^{-8}) = \frac{-1}{4}(u^3 - u^{-3})^2(u^7 + u^{-7})$$ $$(u - u^{-1})(u^8 - u^{-8}) = (u^3 - u^{-3})^2(u^7 + u^{-7})$$ $$u^9 - u^{-7} - u^7 + u^{-9} = (u^6 - 2 + u^{-6})(u^7 + u^{-7})$$ $$u^9 - u^{-7} - u^7 + u^{-9} = u^{13} - 2u^7 + u + u^{-1} - 2u^{-7} + u^{-13}$$ $$u^{22} - u^{6} - u^{20} + u^{4} = u^{26} - 2u^{20} + u^{14} + u^{12} - 2u^{6} + 1$$
All of the exponents are even, so let's simplify by letting $v = u^2$.
$$v^{11} - v^{3} - v^{10} + v^{2} = v^{13} - 2v^{10} + v^{7} + v^{6} - 2v^{3} + 1$$ $$v^{13} - v^{11} - v^{10} + v^{7} + v^{6} - v^{3} - v^{2} + 1 = 0$$
By the Rational Root Theorem, the possible rational roots of this equation are $v = \pm 1$. Sure enough, they both satisfy the polynomial equation. But geometrically, that would require $x$ to be a multiple of $\frac{\pi}{2}$, which doesn't make geometric sense. So let's divide our polynomial by $(v-1)(v+1) = v^2 - 1$ to eliminate these extraneous roots.
$$v^{11} - v^{8} - v^{6} + v^{5} + v^{3} - 1 = 0$$
It turns out that this polynomial also has $v = 1$ (but not $v = -1$) as a root. So divide by $v - 1$ (again).
$$v^{10} + v^9 + v^8 - v^5 + v^2 + v + 1 = 0$$
And at this point, I don't know how to factorize, so I'll cheat and use Wolfram Alpha.
$$(v^4 + v^3 + v^2 + v + 1)(v^6 - v^3 + 1) = 0$$
The first factor turns out not to be particularly useful, but the second one can use the substitution $w = v^3 = u^6$ to make a good old quadratic.
$$w^2 - w + 1 = 0$$ $$w = \frac{1 \pm i \sqrt{3}}{2} = e^{\pm i\pi/3}$$ $$u^6 = e^{\pm i\pi/3}$$ $$u = e^{\pm i\pi/18}$$
So, we get $x = \frac{\pi}{18} = 10°$, as expected.
But I'm not really satisfied with this answer, due to having to solve that high-degree polynomial. Maybe there's an easier way...
On
$\angle(CDB)=5x<\pi\implies x<\dfrac{\pi}{5} $ $\therefore 0<x<\dfrac{\pi}{5}\\ \dfrac{\sin(3x)}{\sin(2x)}=\dfrac{\sin(11x)}{\sin(4x)}\\ \dfrac{\sin(3x)}{ {\sin(2x)}}=\dfrac{\sin(11x)}{2{\sin(2x)}\cos(2x)}\\ 2\sin(3x)\cos(2x)=\sin(11x)\\ \sin(5x)+\sin(x)=\sin(11x)\\ \sin(5x)=\sin(11x)-\sin(x)\\ {\sin(5x)}=2\cos(6x){\sin(5x)}\\ \cos(6x)=\dfrac{1}{2}\\ 6x=\dfrac{\pi}{3}\\ x=\dfrac{\pi}{18}$
Let's do this the other way around.
Assume a triangle $\triangle A'B'C'$ with $\angle B'A'C'=40^{\circ}$ and $A'B'=A'C'$. Let $D'$ be an interior point of $\triangle A'B'C'$ such that $\angle B'A'D'=10^{\circ}$ and $\triangle A'B'D'=20^{\circ}$.
We construct $\triangle A'D'O\cong \triangle A'D'B'$ as shown in the figure above. Then $$\angle B'D'O= \angle B'A'D' + \angle OA'D' + \angle A'B'D' + \angle A'OD'=60^{\circ},$$ implying that $\triangle B'D'O$ is equilateral, so $B'D'=OD'=B'O$. We also obtain $\angle C'B'O=10^{\circ}$.
Since $\angle B'A'O=20^{\circ}=\frac{1}2\angle B'A'C'$, $A'O$ bisects $\angle B'A'C'$. This means $\triangle B'A'O\cong \triangle C'A'O$, so $B'O=C'O$. By symmetry, $\angle B'C'O=\angle C'B'O=10^{\circ}$.
Now we have $B'O=D'O=C'O$, and thus $O$ is the circumcenter of $\triangle B'C'D'$. This implies that $$\angle C'OD'=2\angle C'B'D'=100^{\circ}$$ yielding $$\angle OD'C'=\angle OC'D'=40^{\circ}.$$ Hence $$\angle B'C'D'=\angle OC'D'-\angle B'C'O=30^{\circ}$$ and $$\angle A'C'D'=70^{\circ}-\angle B'C'D'=40^{\circ}.$$
We conclude that $A'\equiv A$, $B'\equiv B$, $C'\equiv C$ and $D'\equiv D$, and $\color{red}{x=10^{\circ}}$ will be our final answer.