In $\triangle ABC$, if $AB=AC$ and internal bisector of angle $B$ meets $AC$ at $D$ such that $BD+AD=BC=4$, then what is $R$ (circumradius)?

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My Approach:- I first drew the diagram and considered $\angle ABD=\angle DBC=\theta$ and as $AB=AC$, $\angle C=2\theta$. Therefore $\angle A=180-4\theta$. Also as $AE$ is the angular bisector and $AB=AC$, then $BE=EC=2.$ Now applying the sine theorem to $\triangle ADB$ and $ \triangle BDC$ gives $$\frac{BD}{\sin {(180-4\theta)}}=\frac{AD}{\sin \theta}$$

$$\frac{BC}{\sin(180-3\theta)}=\frac{BD}{\sin 2\theta}$$
Now we know that $BC=4$ and then solving both the equations by substituting in $BD+AD=4$, we get $$\sin 2\theta .\sin4\theta+\sin2\theta.\sin\theta=\sin3\theta.\sin4\theta$$
Now I have no clue on how to proceed further from here. Though I tried solving the whole equation into one variable ($\sin\theta$), but it's getting very troublesome as power of $4$ occurs. Can anyone please help further or else if there is any alternative method to solving this problem more efficiently or quickly?
Thank You


Indeed you are proceeding correctly. The equation can be solved as follows: $$\sin 2\theta \sin \theta=\sin 4\theta(\sin 3\theta-\sin 2\theta) {\tag 1}$$ Now $\sin 3\theta-\sin 2\theta=2\cos \frac {5\theta}{2} \sin \frac {\theta}{2}$.

Also $\sin \theta=2\cos \frac {\theta}{2} \sin \frac {\theta}{2}$, and $\sin 4\theta=2\sin 2\theta \cos 2\theta$. So, $(1)$ simplifies to: $$\cos \frac {\theta}{2}=2\cos \frac {5\theta}{2}\cos 2\theta {\tag 2}$$ Since $2\cos \frac {5\theta}{2} \cos 2\theta=\cos \frac {9\theta}{2}+\cos \frac {\theta}{2}$, we have, from $(2)$: $$\cos \frac {9\theta}{2}=0$$ This means that $\frac {9\theta}{2}=\frac {\pi}{2}$, hence $\theta=\frac {\pi}{9}$. This means that all angles of triangle are known, and we know $BC=4$. Thus using sine law, $$2R=\frac {BC}{\sin A}$$ is easy to calculate.

  • $\begingroup$ Understood @Ritam_Dasgupta, Thank you very much $\endgroup$
    – PCMSE
    Aug 4 at 6:08
  • 1
    $\begingroup$ I don't think this is quite complete. You can't go from $\cos 9\theta/2 = 0$ to $9\theta/2 = \pi/2$ as there are other solutions, which you have to eliminate. $\endgroup$ Aug 4 at 19:59
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    $\begingroup$ True, I didn't mention it, but it's quite simple really. General solution is $\theta=\frac {(2n+1)\pi}{9}$, and $0<\theta <\frac {\pi}{4}$, since $\angle A>0$, so only this solution is possible $\endgroup$ Aug 5 at 12:23

Yes, there is such a method. Making use of Mathematica command

FullSimplify[Reduce[Sin[2*\[Theta]]*Sin[4*\[Theta]] + 
 Sin[2*\[Theta]]*Sin[\[Theta]] == 
Sin[3*\[Theta]]*Sin[4*\[Theta]] && \[Theta] > 0 && \[Theta] < 
Pi, \[Theta]] // ToRadicals]

, one obtains $$3 \theta =\pi \lor 2 \theta =\pi \lor 9 \theta =\pi \lor 9 \theta =5 \pi \lor 9 \theta =7 \pi.$$

  • $\begingroup$ Sorry, but I haven't learnt this yet. Of course, in future, I will try to learn this. Thank you very much. $\endgroup$
    – PCMSE
    Aug 4 at 6:06

The values are such that equations do not simplify. Nonetheless, here is an alternate approach.

Say $AB = AC = x$ and we know $BC = 4$,

By angle bisector theorem,

$\cfrac{4}{x} = \cfrac{x-AD}{AD} \implies AD = \cfrac{x^2}{4+x}$

Now by angle bisector length formula,

$BD^2 = \cfrac{4x}{(4+x)^2} [(4+x)^2 - x^2] = \cfrac{32x(x+2)}{(4+x)^2}$

Now, $BD + AD = 4 = \cfrac{x^2}{4+x} + \cfrac{\sqrt{32x(x+2)}}{4+x}$

$16+4x-x^2 = \sqrt{32x(x+2)}$

Solving using WolframAlpha, the only valid solution is $x \approx 2.61$

Now to find circumradius, use $R = \cfrac{abc}{4 \triangle} = \cfrac{4 x^2}{8 \sqrt{x^2 - 4}}$

  • 1
    $\begingroup$ Understood @MathLover, thank you very much $\endgroup$
    – PCMSE
    Aug 4 at 14:14
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    $\begingroup$ you are welcome! $\endgroup$
    – Math Lover
    Aug 4 at 14:16

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