Problem Of The Week #449 January 4th 2021

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anemone

MHB POTW Director
Staff member
Here is this week's POTW:

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In a triangle $ABC$, let $\angle A\ge \angle B \ge \angle C$ and suppose that $\sin 4A+\sin 4B+\sin 4C=2(\sin 2A+\sin 2B+\sin 2C)$. Find all possible values of $\cos A$.

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anemone

MHB POTW Director
Staff member
As usual, I will give the community another week's time to attempt at last week's POTW. And I am looking forward to receiving submissions from the members!

anemone

MHB POTW Director
Staff member
No one replied to last two week's POTW. However, you can read the official solution (by other) as below:
Using $A+B+C=\pi$, we have

\begin{align*}\sin 2A+\sin 2B+\sin 2C&=\sin 2A+\sin 2B-\sin (2A+2B)\\&=2\sin (A+B)\cos (A-B)-2\sin(A+B)\cos(A+B)\\&=2\sin(A+B)[\cos(A-B)-\cos(A+B)]\\&=2\sin(A+B)(-2\sin A \sin B)\\&=2\sin C (2\sin A\sin B)\\&=4\sin A \sin B\sin C\end{align*}

Replace $A,\,B$ and $C$ in the above calculation by $2A,\,2B$ and $2C$ to get

$\sin 4A+\sin 4B+\sin 4C=-4\sin 2A\sin 2B\sin 2C$

Using these two inequalities, the given relation is equivalent to

$-4cos A\cos B\cos C=1$

Since the product of cosines in $-4\cos A\cos B\cos C=1$ is negative, and $A\ge B\ge C$, we must have $A>\dfrac{\pi}{2}>2C>0$.

Using $A+B+C=\pi$, from $-4\cos A\cos B\cos C=1$, we get

$4\cos A \cos(A+C)\cos C=1$ and so

$4\cos A(\cos A \cos C-\sin A \sin C)\cos C=1$

Divide both sides by $\cos^2 C$ to get

$4\cos^2 A-4\sin A\cos A\tan C=\dfrac{1}{\cos^2 C}$

Let $x=\cos A$ and $t=\tan^2 C$.

From $A>\dfrac{\pi}{2}>2C>0$, it follows that $-1<x<0$ and $0<t<1$.

Rewrite $4\cos^2 A-4\sin A\cos A\tan C=\dfrac{1}{\cos^2 C}$ using this notation and rearrage to get

$4x^2-(1+t)=4x\sqrt{1-x^2}\sqrt{t}$

Square both dies and move all the terms to one side to get

$16(1+t)x^4-8(1+3t)x^2+(t+1)^2=0$

Applying the quadratic formula, we get

$x^2=\dfrac{1+3t\pm\sqrt{3t+6t^2-t^3}}{4(1+t)}$

Since $-1<x<0$ and $0<t<1$, $4x^2-(1+t)<0$ whenever $4x^2-(1+t)=4x\sqrt{1-x^2}\sqrt{t}$ holds.

For the solutions obtained from the quadratic formula we have

$4x^2-(1+t)=\dfrac{(t-t^2)\pm \sqrt{3t+6t^2-t^3}}{1+t}$ and since $t-t^2>0$ it is clear that when $x^2=\dfrac{1+3t\pm\sqrt{3t+6t^2-t^3}}{4(1+t)}$, $4x^2-(1+t)=4x\sqrt{1-x^2}\sqrt{t}$ is not satisfied.

Therefore, since $x<0$, we must have

$x=-\dfrac{1}{2}\sqrt{\dfrac{1+3t-\sqrt{3t+6t^2-t^3}}{1+t}}$

We want to find the range of values of $x$ for $t\in(0,\,1)$. Implicitly differentiate $x^2=\dfrac{1+3t\pm\sqrt{3t+6t^2-t^3}}{4(1+t)}$ to get

$x\dfrac{dx}{dt}=\dfrac{4\sqrt{3t+6t^2-t^3}+t^3+3t^2-9t-3}{16(1+t)^2\sqrt{3t+6t^2-t^3}}$

There is no $t\in(0,\,1)$ for which either $x=0$ or $16(1+t)^2\sqrt{3t+6t^2-t^3}=0$, so we conclude that the critical points of $x$ satisfy

$4\sqrt{3t+6t^2-t^3}=-t^3-3t^2+9t+3\\(3+9t-3t^2-t^3)^2-16(3t+6t^2-t^3)=0\\t^6+6t^5-9t^4-44t^3-33t^2+6t+9=0\\(t-3)(t+1)^3(t^2+6t-30$

The only critical point in the range $[0,\,1]$ is $t=2\sqrt{3}-2$. The corresponding value of $x$, obtained after a tedious but straightforward calculation is $\dfrac{1=\sqrt{3}}{2}$. From

$x=-\dfrac{1}{2}\sqrt{\dfrac{1+3t-\sqrt{3t+6t^2-t^3}}{1+t}}$,

we easily evaluate

$\displaystyle \lim_{t\rightarrow 0^+} x=-\dfrac{1}{2}$ and

$\displaystyle \lim_{t\rightarrow 1^-} x=-\dfrac{\sqrt{2-\sqrt{2}}}{2}$

allowing us to conclude that for $t\in(0,\,1)$, we have

$-\dfrac{1}{2}<x\le \dfrac{1-\sqrt{3}}{2}$

Finally, we check that for each $x$ in this interval there is a corresponding triangle whose angles $A,\,B$ and $C$ satisfy the given relation. Suppose $x_0$ is such that $\dfrac{1}{2}<x_0\le \dfrac{1-\sqrt{3}}{2}$. Let $A=\cos ^{-1}x_0$. Since $\cos^{-1}$ is a decreasing function we have $A<\cos^{-1}(-0.5)=\dfrac{2\pi}{3}$.

By the intermediate value theorem, since $x$ is continuous on $0,\,2\sqrt{3}-3]$, there exists a $t+0$ in this interval such that $x_0=x(t_0)$.

Let $C=\tan^{-1}\sqrt{t_0}$. Note that $\sqrt{t_0}\le \sqrt{2\sqrt{3}-3}<\sqrt{3}$, where $C<\dfrac{\pi}{3}$.

Let $B=\pi-A-C$, the earlier comments about the ranges for $A$ and $C$ imply $B>0$.

We claim that a triangle with angles $A,\,B$ and $C$ satisfies the relation given in the problem. From the construction, $\cos C=(1+\tan^2 C)^{-0.5}=(1+t_0)^{-0.5}$ and $\cos A=x_0$. Moreover, $x_0$ and $t$ satisfy equation $4x^2-(1+t)=4x\sqrt{1-x^2}\sqrt{t}$. We calculate, using trigonometric equalities to evaluate $\sin A$ and $\sin C$.

\begin{align*}\cos B&=\cos (\pi-(A+C))\\&=-\cos (A+C)\\&=-\cos A\cos C+\sin A \sin C\\&=-x_0\sqrt{\dfrac{1}{1+t_0}}+\sqrt{1-x_0^2}\cdot \sqrt{1-\dfrac{1}{1+t_0}}\\&=\sqrt{\dfrac{1}{1+t_0}}(-x_0+\sqrt{1-x_0^2}\cdot \sqrt{t_0})\end{align*}

Since $x_0\ne 0$, we rearrange $4x^2-(1+t)=4x\sqrt{1-x^2}\sqrt{t}$ to get $\sqrt{1-x_0^2}\cdot \sqrt{t_0}=x_0-\dfrac{1+t_0}{4x_0}$.

Hence $\cos B=\sqrt{\dfrac{1}{1+t_0}}\cdot \dfrac{1+t_0}{-4x_0}=\dfrac{\cos C(1+t_0)}{-4\cos A}$.

It follows that $-4\cos A\cos B\cos C=1$, so $A,\,B,\,C$ satisfy $-4cos A\cos B\cos C=1$, which is equivalent to the equality in the question.

Therefore, we conclude that the possible range of value for $\cos A$ is given by $-\dfrac{1}{2}<\cos A\le \dfrac{1-\sqrt{3}}{2}$.

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