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The formula for the circumradius $r$ of a triangle $ABC$ tells me that $r={abc\over{}4\triangle}$, where the lengths of the sides are $a$, $b$, $c$.

I'm not sure, but I occasionaly got wrong values. They might have been calculation mistakes, but then I got fixated on deriving the formula for myself.

So, I took a triangle $ABC$, circumcenter $R$.

I know that:

  • $RA=RB=RC$
  • Altitudes from $R$ to sides bisect those sides.

Side $AB$ was divided into two parts each of length $c$, $BC$ into two of $a$, and $AC$ into two of $b$. Let the altitudes to $BC$ be of length $h_1$, to $AC$ be of length $h_2$, and to $AB$ of length $h_3$.

So, using the Pythagorean Theorem,

$\begin{align}r^2&=a^2+h_1^2\\r^2&=b^2+h_2^2\\r^2&=c^2+h_3^2\end{align}$

Since I know $a,b,c$, I have four variables, namely $r,h_1,h_2,h_3$. But since I have only three equations, I am unable to solve. I have tried many times, yet I cannot find any other relations. So my primary problem is to find the fourth equation.

Please help.

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  • $\begingroup$ it would be a lot easier to use $a=2r \sin A$ and $\frac{1}{2}b c \sin A =\Delta $ $\endgroup$ – Lozenges Nov 22 '17 at 15:11
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We can get this formula by the following way.

Let $\angle C$ be an acute angle, $BD$ be an altitude of $\Delta ABC$ and $BE$ be a diameter of the circumcircle.

Thus, $\Delta ABE\sim\Delta DBC$ and $$\frac{c}{h_b}=\frac{2r}{a},$$ which gives $$r=\frac{ac}{2h_b}=\frac{ac}{2\cdot\frac{2\Delta}{b}}=\frac{abc}{4\Delta}.$$

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  • $\begingroup$ How is $\Delta ABD\sim\Delta BDC$? Angle D is right, angle B is not. $\endgroup$ – DynamoBlaze Nov 22 '17 at 15:15
  • $\begingroup$ Because $\measuredangle E=\measuredangle C$. $\endgroup$ – Michael Rozenberg Nov 22 '17 at 15:17
  • $\begingroup$ Did you mean $\Delta ABE\sim\Delta BDC$, instead of $\Delta ABD\sim\Delta BDC$? $\endgroup$ – DynamoBlaze Nov 22 '17 at 15:18
  • $\begingroup$ It was typo. I fixed. See now. $\endgroup$ – Michael Rozenberg Nov 22 '17 at 15:19
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You can use the fact that the area of the triangle $ABC$ will be the sum of areas of three isosceles triangles with bases $a$, $b$, $c$ and heights $h_1$, $h_2$, $h_3$. So $A_{ABC}=\frac{1}{2}(ah_1+bh_2+ch_3)$. You can use Heron's formula to find the area of the triangle $ABC$.

Also, if you want to use the law of sines, $\frac{AB}{\sin C}=\frac{BC}{\sin A}=\frac{AC}{\sin B}=2r$ and you can express those sines using $h_1$, $h_2$, $h_3$ along with the sides and the radius to get additional equations.

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enter image description here

The marked angles are equal by the Inscribed Angle Theorem.

Note that by similar triangles, $$ \frac hb=\frac{c/2}r\tag1 $$ Thus, the area of the triangle is $$ A=\frac{ah}2=\frac{abc}{4r}\tag2 $$ Therefore, the circumradius is $$ r=\frac{abc}{4A}\tag3 $$

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  • $\begingroup$ This may just add a bit of detail and a diagram to Michael Rozenberg's answer, but it seemed a useful addition. $\endgroup$ – robjohn Apr 5 at 14:25

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