All the perpendicular bisectors of the sides of a triangle intersect at one point — the circumcenter. So, why do we typically find all three perpendicular bisectors when trying to circumscribe a triangle in a circle? Wouldn't two perpendicular bisectors be sufficient to find the radius of the circle since the third perpendicular bisector would intersect in the same place anyway?
Why do you need three perpendicular bisectors to circumscribe a triangle
geometry
Related Solutions
angle bisector $AI$ cut the circumcircle of $\triangle ABC$ at $D$
$\angle DBI=\angle DBC+\angle IBC=\angle DAB+\angle ABI=\angle BID$ and then $DB=DI$
Likewise, $DC=DI$ and then $DB=BI=DC$
$I_{A}C$ bisect $\angle BCT$ $\Longrightarrow$ $\angle ICI_{A}=90^{\circ}$
thus, $DI=DC=DI_{A}$
the perpendicular bisector of $BC$ cut the circumcircle of $\triangle ABC$ at $M$
$\triangle SAI_{A}$ is similar to $\triangle BMD$
power of $I_{A}$ with respect to the circumcircle of $\triangle ABC$ is $OI_{A}^{2}-R^{2}$
Also, it is $I_{A}D\cdot I_{A}A$
$\triangle SAI_{A}\sim\triangle BMD$ $\Longrightarrow$ $\dfrac{MD}{BD}=\dfrac{I_{A}A}{SI_{A}}$
$\Longrightarrow$ $2Rr_{A}=BD\cdot I_{A}A=DI_{A}\cdot I_{A}A$
hence, $2Rr_{A}=DI_{A}\cdot I_{A}A=OI_{A}^{2}-R^{2}$
To incorporate all possible cases it is convenient to allow the values $p_i$ be signed, negative sign meaning that the side $i$ lies against the obtuse angle. There can be only one negative $p_i$ and it has to be the least of three by the absolute value. If all three $p_i$ are positive or the least by the absolute value is negative and two other are positive the solution always exists and is unique.
The finding of $a,b,c$ can be reduced to finding the circumradius $r$. For this one should solve the equation: $$ \arccos\frac{p_a}r+\arccos\frac{p_b}r+\arccos\frac{p_c}r=\pi.\tag1 $$ Using the formula $$ \cos(x+y+z)=\cos x\cos y\cos z-\cos x\sin y\sin z-\sin x\cos y\sin z-\sin x\sin y\cos z $$ one can after some straightforward algebra reduce the equation $(1)$ to: $$ r^3-3pr-2q=0.\tag2 $$ with $$ p=\frac{p_a^2+p_b^2+p_c^2}3,\quad q=p_a p_b p_c.\tag3 $$
This cubic equation has depressed form and can be easily solved. Since due to AM-GM inequality we have in general $$ -p^3+q^2\le0 $$ the equation has three real roots and we need the largest one.
The simplest way to represent the required soluton of $(2)$ is to use the trigonometric form: $$ r=2p^{1/2}\cos\left[\frac13\arccos\left(\frac q{p^{3/2}}\right)\right].\tag4 $$
Best Answer
First, even though this isn't what you asked, I wanted to explain why the 3 perpendicular bisectors are concurrent in the first place, and why they intersect at the circumcenter.
Basically, in order to circumscribe a triangle using a circle, you need to find the point that is equidistant from each of the vertices in the triangle. A perpendicular bisector of a segment AB is actually the set of all points that are equidistant from A and B. You can see this by picking an arbitrary point C on the line, and the intersection of the perpendicular bisector of AB with AB D. It is easy to see that AD=DB, CD=CD, and m<ADC=<BDC=90ยบ. Therefore, triangles ADC and BDC are congruent by SAS, so AC=BC.
Now let's focus on the triangle in question. Call it XYZ. Call line z the perpendicular bisector of XY, y the bisector of XZ, and x the bisector of YZ. The intersection of x and z is the point equidistant from X, Y, and Z, and since two lines can only intersect at one point, there is only one point with this property, This means that, since x and y also intersect at a point with this property, and the only point with that property is the intersection of x and z, we know that x, y, and z must be concurrent at the point equidistant from X, Y, and Z, called the circumcenter of triangle XYZ.
So, yes, you can find the circumcenter using only two perpendicular bisectors, but using 3 is usually a good thing to do since if they don't concur at the same point you know you made a mistake :)