# Radius of the circumscribed circle of an isosceles triangle

An isosceles triangle $$ABC$$ is given $$(AC=BC).$$ The perimeter of $$\triangle ABC$$ is $$2p$$, and the base angle is $$\alpha.$$ Find the radius of the circumscribed circle $$R$$.

$$R=\frac{p}{2\sin\alpha(1+\cos\alpha)}$$

Let $$CD=2R.$$ The triangle $$BCD$$ is a right triangle and we have $$\angle BAC=\angle ABC=\angle BDC=\alpha.$$

I am not sure how to approach the problem. It's really hard for me to solve problems like this. Can you give me a hint and some thoughts on the problem?

• Please see how to ask a good question. Sep 5, 2020 at 9:37
• I added what I tried. As I said in the post I don't know how to approach problems like that. What do you want me to do? Sep 5, 2020 at 9:40
• Are you learning geometry or trigonometry right now? If so, what topics have you learnt so far? If you can, adding a diagram would really help. Sep 5, 2020 at 9:42
• I added a diagram. :) Sep 5, 2020 at 9:49
• Fantastic, thanks for adding one. Sep 5, 2020 at 9:49

Another simple approach. Let $$x=AC=BC$$. Then

$$2p=AC+BC+2AH\\=2x+2x\cos\alpha$$

and

$$R=\frac 12 CD=\frac 12 \frac{BC}{ \sin \alpha} = \frac{x}{2 \sin \alpha}$$

Now you can complete the solution by a simple substitution.

• Thank you for the response! I am not sure I understand what to do next. Sep 5, 2020 at 10:39
• Simply solve the first equation for $x$ and then substitute the result in the second equation. Sep 5, 2020 at 11:15

Hint: Use following formula:

$$R=\frac{p}{4\cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\beta}{2}\right)\cos\left(\frac{\gamma}{2}\right)}$$

Where $$\alpha$$, $$\beta$$ and $$\gamma$$ are angles on vertices A, B and C respectively. $$\alpha=\beta$$ , therefore we have:

$$R=\frac{p}{4\cos^2\left(\frac{\alpha}{2}\right)\cos\left(\frac{\gamma}{2}\right)}$$

And also:

$$2\alpha+\gamma=\pi$$

$$\implies\frac{\gamma}{2}=\frac{\pi}{2}-\frac{\alpha}{4}$$

Finally:

$$R=\frac{p}{4\cos^2\left(\frac{\alpha}{2}\right)\sin\left(\frac{\alpha}{4}\right)}$$

The diameter is $$CD = 2R = \sqrt{BD^2 + BC^2}$$ by the Pythagorean theorem, since $$\angle CBD$$ is inscribed in a semicircle, thus is a right angle.

Now use the trigonometric properties to deduce that $$BH = BD \sin \alpha,$$ and $$BH = BC \cos \alpha.$$ We also have $$BH + BC = p,$$ because this is half the perimeter of $$\triangle ABC$$. Now all that is left is to eliminate $$BH$$, $$BD$$, and $$BC$$ from these four equations.

Using two known general expressions for the area of $$\triangle ABC$$

\begin{align} S&=\rho r \tag{1}\label{1} ,\\ S& =2\,R^2\sin\alpha\sin\beta\sin\gamma =4\,R^2\sin^3\alpha\cos\alpha \tag{2}\label{2} , \end{align}

and the expression for the inradius of $$\triangle ABC$$ in terms of its semiperimeter $$\rho$$,

\begin{align} r&= \rho\tan\tfrac\alpha2\tan\tfrac\beta2\tan\tfrac\gamma2 = \rho\tan^2\tfrac\alpha2\cot\alpha \tag{3}\label{3} , \end{align} we can find that

\begin{align} R&= \frac1{2\sin\alpha}\,\sqrt{ \frac{\rho r}{\sin\alpha\cos\alpha} } \tag{4}\label{4} ,\\ R&= \frac1{2\sin\alpha}\,\sqrt{ \frac{\rho^2 \tan^2\tfrac\alpha2\cot\alpha}{\sin\alpha\cos\alpha} } =\dots= \frac\rho{2\sin\alpha+\sin2\alpha} \tag{5}\label{5} . \end{align}