Circumcenter

The circumcenter of a polygon is the center of the circle that contains all the vertices of the polygon, if such a circle exists. For a triangle, it always has a unique circumcenter and thus unique circumcircle.

This wiki page is an overview of the properties of the circumcenter of a triangle, which are applied to different scenarios like Euclidean geometry. After reading this page, you should have a substantial understanding of circumcenter and all the related ingredients, and will be able to solve the problems proposed on the notions of circumcenters and circumcircles.

Let's look into the basic properties of circumcircles and some relevant examples. We can derive angle-related properties using the properties of angle in a circle (inscribed angle). Writing down the properties, however, may be a bit tricky since the circumcenter $O$ can lie either outside or inside triangle $ABC$.

Properties

• $O$ is the intersection of perpendicular bisectors of the sides of triangle $ABC$.

• Connecting the circumcenter $O$ to the vertices of triangle $ABC$ gives three isosceles triangles $OAB, OBC, OCA$.

• $R= \dfrac{a}{2 \sin A } = \dfrac{abc}{ 4S}$ follows from the extended sine rule, where $S$ denotes the area of the triangle $ABC.$

• Location of circumcenter differs for the acute, obtuse, and right-angled triangles. This can be deduced from the central angle property:

  • If $\angle B$ is acute, then $\angle BOC=2\angle A$.
    
  • If $\angle B$ is right, then $O$ lies on the midpoint of $AC$.
    
  • If $\angle B$ is obtuse, then $O$ lies on the opposite side of $AC$ from $B$ and $\angle BOC=2\angle A$.

• Moreover, the alternate segment theorem states that the angle formed by the tangent to the circle and one of the sides of the triangle is equal to the angle between the other two sides:

In the next section we will learn how these applications can be used for solving problems.

Now that we've gone through the properties possessed by a circumcircle and its circumcenter, we will apply them to work out some examples and problems which should make you more confident in their usage.

What is the circumcenter of triangle $ABC$ with vertices $A=(1, 4), B=(-2, 3), C=(5, 2)$?

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By definition, a circumcenter is the center of the circle in which a triangle is inscribed. For this problem, let $O=(a, b)$ be the circumcenter of $\triangle ABC.$ Then, since the distances to $O$ from the vertices are all equal, we have $$\overline{AO} = \overline{BO} = \overline{CO} .$$ From the first equality, we have $$\begin{aligned} \overline{AO} ^2&= \overline{BO} ^2\\ (a-1)^2+(b-4)^2&=(a+2)^2+(b-3)^2\\ -2a+1-8b+16&=4a+4-6b+9\\ 3a+b&=2. &&\qquad (1) \end{aligned}$$ Similarly, from the second equality, we have $$\begin{aligned} \overline{BO}^2&=\overline{CO}^2\\ (a+2)^2+(b-3)^2&=(a-5)^2+(b-2)^2\\ 4a+4-6b+9&=-10a+25-4b+4\\ 7a-b&=8. &&\qquad (2) \end{aligned}$$ Taking $(1)+(2)$ gives $a=1,$ which in turn gives $b=-1.$

Therefore, the circumcenter of triangle $ABC$ is $O=(1, -1).$ $_\square$

We can use other properties of the circumcircle to help us find the circumcenter to solve this problem. For example, you can use the fact that it is the intersection of the perpendicular bisectors.

What is the radius of the circumcircle for the equilateral triangle with side length 2?

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Let $O$ be the circumcenter, and $D$ the foot of the perpendicular from $A$ to $\overline{BC}$. Since $\overline{OB} =\overline{OC}$, this tells us that $O$ lies on the perpendicular bisector of $\overline{BC}$, and thus $O$ lies on $\overline{AD}$. In this diagram, we have $\overline{OA} = \overline{OB} = \overline{OC} = R$, the radius of the circumcircle. Since $\overline{BD}=\overline{CD}=1,$ the height of triangle $ABC$ is $\overline{AD} = \sqrt{3}$. Then, from the right triangle $OBD$, we obtain $\overline{OD}^2 + \overline{BD}^2 = \overline{OB}^2$, or $\big(\sqrt{3} - R\big)^2 + 1^2 = R^2$. Expanding this and solving for $R,$ we obtain $$3 - 2 \sqrt{3} R + R^2 + 1 = R^2 \implies 4 = 2 \sqrt{3} R \implies R = \frac{2 \sqrt{3} }{3}. \ _\square$$

We have worked out the above examples so that you can get familiar with using the properties above in problem solving. Now it's your turn to try a couple of problems on your own and make the properties worth learning.

Find the area of the circumcircle of an isosceles triangle $ABC$ above, where $a=b$ and $\angle A=15^\circ,$ provided that the perimeter of the triangle is $25 \text{ cm}.$

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Let $R$ be the circumradius of $\triangle ABC.$ Then, from the extended sine rule, we have

$$\begin{aligned} \dfrac{a}{\sin A} =\dfrac{b}{\sin B}=\dfrac{c}{\sin C} &=2R\\ \Rightarrow R&=\dfrac{a}{2 \sin A}. \end{aligned}$$

Since we are given $a=b$ and $a+b+c=25,$ it follows that $c=25-2a,$ which implies

$$\dfrac{a}{\sin A}=\dfrac{c}{\sin C}=\dfrac{25-2a}{\sin C}.$$

Since it is given that $\angle A=15^\circ$ and $\angle C=150^\circ,$ solving further gives $a=\dfrac{50 \sin 15^\circ}{1+4 \sin 15^\circ}$ and $R=\dfrac{25}{1+4 \sin 15^\circ}$.

Hence the area of the circumcircle is $\pi R^2=\pi \left( \dfrac{25}{1+4 \sin 15^\circ} \right)^2=474.005.$ $_\square$

Find the area of rhombus $ABCD$ given that the radii of the circumcircles of triangles $ABD$ and $ACD$ are $12.5$ and $25,$ respectively.

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The radius of the circumcircle of triangle $ABD$ is $\left( \dfrac{d_1}{2 \sin A} \right)=12.5. \qquad (1)$
The radius of the circumcircle of triangle $ACD$ is $\left( \dfrac{d_2}{2 \sin (180^\circ - A)} \right)=25. \qquad (2)$
Taking $(2)\div (1),$ we get $\frac{d_2}{d_1}=2 \implies d_2=2d_1. \qquad (3)$

The area of the rhombus is $\frac{d_1d_2}{2}$. Using the formula for the radius of the circumcircle of a triangle, we can write the area of $\triangle ABC$ as $$A = \dfrac{abc}{4R} = \dfrac{s^2d_2}{(4)(25)},$$

where the side length of the rhombus $s$ can be written as $\frac{1}{2}\sqrt{d_1^2 + d_2^2}$. Since twice the area of this triangle is equal to the area of the rhombus, we have

$$\frac{d_1d_2}{2} = \frac{\left(d_1^2 + d_2^2\right)d_2}{(8)(25)} \implies 20d_1 = d_1^2. \qquad (\text{since } d_2 = 2d_1)$$

This means $d_1 = 20$ and $d_2 = 40$. Therefore, the area of the rhombus is 400. $_\square$

In this section, we will try problems that can be greatly simplified by applying properties of the circumcenter, even though the problems themselves don't directly involve the circumcenter. Since the circumcenter is a rich structure that interrelates angles and lengths, using it correctly in a problem (e.g. International Mathematical Olympiad, or IMO) can be very powerful. For this reason, it is important to know how to spot circumcenters and the appropriate moment to use them.

$O$ is the circumcenter of $ABC$ if and only if

1. $AO=BO=CO$;
2. $BO=CO$ and $\angle BOC=2\angle A$ when $\angle A$ is acute and $A,O$ are on the same side of $BC$;
3. $BO=CO$ and $\angle BOC=2(180-\angle A)$ when $\angle A$ is obtuse and $A,O$ are on opposite sides of $BC$.

Let's see some of these criteria in action.

Quadrilateral $ABCD$ satisfies $\angle A=76^{\circ}, \angle B=72^{\circ}, \angle C=142^{\circ}$, and $AD=AB$. Find $\angle ACB-\angle ACD.$

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Since $AD=AB$ and the angles of the quadrilateral are given, it does not hurt to see if $A$ is a circumcenter.

By the definition of a quadrilateral we know $A,C$ are on opposite sides of $BD$. In addition,

$$2(180^\circ-\angle BCD)=2(180^\circ-142^\circ)=76^{\circ}=\angle BAD.$$

Then, from criterion number 3 above, we know that $A$ is the circumcenter of $\triangle BCD$.

Therefore, $\angle ACB=\angle ABC=72^{\circ}, \angle ACD=\angle ADC=360^\circ-76^\circ-72^\circ-142^\circ=70^{\circ}$, and our answer is $72^\circ-70^\circ=2^{\circ}.$ $_\square$

Note: As one can see, realizing that $A$ is the circumcenter of $\triangle BCD$ immediately made the problem clear. It is fine to say that the circumcenter is the hidden configuration in this problem, and revealing it is the key. Now imagine approaching this with trigonometry. Although considering it is natural, using trig here will definitely be a pain.

(Indian National Mathematical Olympiad 2015)
Let $ABC$ be a right-angled triangle with $\angle B = 90^{\circ}.$ Let $BD$ be the altitude from $B$ on to $AC.$ Let $P, Q$ and $I$ be the incenters of triangles $ABD, CBD$ and $ABC,$ respectively. Show that the circumcenter of triangle $PIQ$ lies on the hypotenuse $AC$.

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Suppose $O$ is the circumcenter of triangle $PIQ$. Since $\angle PIQ=\angle AIC=135^{\circ}$, by the angle properties of cirumcenter $\angle POQ=2(180-135)=90^{\circ}$. Note that $\angle PDQ=90^{\circ}$. Hence $D,O,P,Q$ are concyclic.

Since $PO=QO$, $DO$ externally bisects $\angle PDQ$. However, $AC$ is also the external bisector of $\angle PDQ$. In conclusion, $O$ must lie on $AC.$ $_\square$

The following is an advanced example that requires you to recognize the circumcenter and utilize it to perform crucial angle chasing.

$\triangle ABC$ is a right triangle with $\angle B=90^\circ$. $D$ is a point inside $\triangle ABC$ such that $\overline{AD}=\overline{AB},$ and $E$ is on $\overline{AC}$ satisfying $\overline{DE}\perp \overline{BD}$. Also, $F$ is the point where the circumcircle of $\triangle CDE$ intersects $\overline{BC}.$

Prove that $\overline{BE}=\overline{EF}$.

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We want to show that $\angle EBF = \angle EFB$. Since $\angle EFB = 180 ^ \circ - \angle EFC = 180^ \circ - \angle EDC$, it suffices to show that $\angle EBC + \angle EDC = 180 ^ \circ$.

We almost want to say they are opposite angles of a cyclic quad, but $B, D$ lie on the same side of $EC$, so that won't be possible. This motivates the construction of $D'$ as the reflection of $D$ in the line $EC$, and it suffices to show that $BED'C$ is a cyclic quad. There are many different approaches that we could use, and we will show that $\angle ECB = \angle ED'B$. We pick this angle because $\angle ACB$ is fixed in this question, while $D, E, D'$ are variable points.

Now, observe that $\overline{AB} = \overline{AD} = \overline{AD'},$ so $A$ is the circumcenter of $\triangle B D D'$. This means that $\angle BD' D = \frac{1}{2} \angle BAD$. This motivates constructing $M$ as the midpoint of $BD$, so that we get $\angle BD'D = \angle BAM .$ Hence, it suffices to show that $\angle ED'B + \angle BD'D = \angle ACB + \angle BAM$.

Observe that $AM \parallel ED$ since they are both perpendicular to $BD$. We then have $\angle ACB + \angle BAM = 90 ^ \circ - \angle MAC = 90^ \circ - DEC = \angle EDD'$. This is equal to $\angle ED'D$ due to isosceles triangle $ED'D$. Hence we are done. $_\square$

Our first problem can be solved by constructing similar triangles and proceeded by angle chasing. See below for a more direct proof that arose from recognizing concyclic points.

Let $M$ be the midpoint of side $AC$ of acute triangle $ABC$ with $\overline{AB}>\overline{BC},$ and let $\Omega$ be the circumcircle of $\triangle ABC$. The tangents to $\Omega$ at points $A$ and $C$ meet at $P$, and $BP$ and $AC$ intersect at $S$. Also, let $AD$ be the altitude of $\triangle ABP,$ and $\omega$ the circumcircle of $\triangle CSD$. Given that $\omega$ and $\Omega$ intersect at $K\not= C,$ prove that $\angle CKM=90^{\circ} .$

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Since $\overline{AP}=\overline{CP}$, we have $\angle AMP=90^{\circ}=\angle ADP,$ which implies $A,P,D,M$ are concyclic. Since $C,K,D,S$ are given to be concyclic, by the alternate segment theorem $\angle KAP=\angle ACK=\angle KDP \implies K\in \odot (APDM)$. Applying the properties of cyclic quadrilaterals and alternate segment theorem again, we have

$$\begin{aligned}\angle MKP=180^{\circ}-\angle CAP=\angle AKC \implies \angle CKM&=\angle AKC-\angle AKM\\&=\angle MKP-\angle AKM\\&=\angle AKP=90^{\circ}. \ _\square\end{aligned}$$

Believe it or not, a lot of IMO problems can be solved simply by some clever constructions of cyclic points. Here's an example from 2014:

[IMO 2014]
Convex quadrilateral $ABCD$ has $\angle{ABC}=\angle{CDA}=90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside $\triangle{SCT}$ and

$$\begin{array}{c}&\angle{CHS}-\angle{CSB}=90^{\circ}, &\angle{THC}-\angle{DTC} = 90^{\circ}.\end{array}$$

Prove that line $BD$ is tangent to the circumcircle of $\triangle{TSH}$.

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First of all, points $S,T$ are defined in an unconventional manner, so let's try to tackle them first. Rearranging the angle equality gives

$$\angle CHS=90+\angle CSB=180-\angle SCB.$$

Seeing this reminds us of opposite angles in a cyclic quadrilateral summing to $180^{\circ},$ and is enough to motivate us to reflect $C$ over $AB,AD$ to obtain $E,F$ with cyclic $CHSE, CHTF,$ respectively.

Now there are several ways to show that the problem is equivalent to proving $CH\perp ST$, one of which will be presented at the end.

Project $C$ onto $SH,TH$ to get $J,K,$ respectively. Right away we observe cyclic $CJSB,CKTD$. Notice $\angle CHK=\angle TFC=\angle TCF=\angle TKD$, and since $CKH$ is a right triangle at $K$, $KD$ bisects $CH$. Similarly we can show that $BJ$ bisects $CH$, implying $BJ,DK$ intersection on the midpoint of $HC$, denoted by $M$, or the circumcenter of $CJHK$. It is well known that the reflection of $C$ over the perpendicular bisector of $BD$ lies on $AH$, and hence $M$ lies on perpendicular bisector of $BD$, which is sufficient to conclude that $BDJK$ is an isosceles trapezoid, a cyclic quadrilateral.

Consider the radical axis of $\odot CJSB,\odot CKTD$, which is perpendicular to $ST.$ We then know $BJ\cap DK=M$ lies on it by radical axis theorem. Consequently line $CMH$ is the radical axis, establishing that $CH\perp ST$. Hence

$$\angle TSH=\angle 90^\circ-\angle JHC=\angle HCJ=\angle HKJ=\angle THD.$$

By the alternate segment theorem, converse $BD$ is tangent to the circumcircle of $TSH,$ and we are done. $_\square$

In the next example, a solution with a clear intention to create a cyclic figure is showcased. The technique is very similar to the one used in the proof example in the last section.

[2011 IMO SL6]
Let $ABC$ be a triangle with $\overline{AB}=\overline{AC}$ and let $D$ be the midpoint of $AC$. The angle bisector of $\angle BAC$ intersects the circle through $D,B$ and $C$ at the point $E$ inside the triangle $ABC$. The line $BD$ intersects the circle through $A,E$ and $B$ in two points $B$ and $F$. The lines $AF$ and $BE$ meet at a point $I$, and the lines $CI$ and $BD$ meet at a point $K$. Show that $I$ is the incentre of triangle $KAB.$

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From cyclic and symmetry, we can show $\angle EBD=\angle ECD=\angle ABE,$ which implies $BI$ bisects $\angle KBA$. By the properties of incenter, $I$ is the incenter of $KAB\iff \angle AIC=\angle ABI+90^{\circ}$. Note that

$$\angle AFD=180^{\circ}-\angle BAE=\frac {\angle BEC}{2}=\frac {\angle BDC}{2},$$

which implies

$$\angle DAF=\angle BDC-\angle AFD=\angle AFD\implies AD=DF=DC\implies \angle AFC=90^{\circ}.$$

So now it suffices to show $\angle ABI=\angle AIC-90^{\circ}=\angle ICF$.

Reflect $B$ over $AF$ to obtain $B'$. Now $C,B'$ are on the same side of $IF$, so we just need to show $I,C,B',F$ are concyclic, which is equivalent to $\angle IB'C=\angle IFC=90^{\circ}$. Since $IB'$ internally bisects $\angle FB'A$ by symmetry, we want to prove $B'C$ externally bisects $\angle FB'A$, or $180^{\circ}-\angle FB'C=\angle AB'C=\angle ACB'$ since $AB'=AB=AC$. The last equivalence amounts to proving $FB'||AC$, which is true by $180^{\circ}-\angle AFB'=\angle AFD=\angle CAF.$ $_\square$

Intersection of Circles

Two circles in a plane intersect in zero, one, two, or infinitely many points. The latter case occurs only in the case of two identical circles. The first case (zero points of intersection) occurs whenever the distance between the centers of the circles is greater than the sum of the radii, or the distance between the centers is less than the absolute value of the difference in their radii. Here are all the cases, explained graphically:

Incenter and Incircle of Triangles

An incircle is an inscribed circle of a polygon, i.e. a circle at which each side of the polygon is a tangent. The center $I$ of the incircle is called the Incenter and the radius $r$ of the circle is called the inradius. The incenter is the point of concurrence (where they cross) of the triangle’s angle bisectors.

While an incircle does not necessarily exist for arbitrary polygons, it exists and is moreover unique for triangles and regular polygons.