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 can lie either outside or inside triangle .
is the intersection of perpendicular bisectors of the sides of triangle .
Connecting the circumcenter to the vertices of triangle gives three isosceles triangles .
follows from the extended sine rule.
Location of circumcenter differs for the acute, obtuse, and right-angled triangles. This can be deduced from the central angle property:
- If is acute, then .
- If is right, then lies on the midpoint of .
- If is obtuse, then lies on the opposite side of from and .
- If is acute, then .
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 with vertices ?
By definition, a circumcenter is the center of the circle in which a triangle is inscribed. For this problem, let be the circumcenter of Then, since the distances to from the vertices are all equal, we have From the first equality, we have Similarly, from the second equality, we have Taking gives which in turn gives
Therefore, the circumcenter of triangle is
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?
Let be the circumcenter, and the foot of the perpendicular from to . Since , this tells us that lies on the perpendicular bisector of , and thus lies on . In this diagram, we have , the radius of the circumcircle. Since the height of triangle is . Then, from the right triangle , we obtain , or . Expanding this and solving for we obtain
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 above, where and provided that the perimeter of the triangle is
Let be the circumradius of Then, from the extended sine rule, we have
Since we are given and it follows that which implies
Since it is given that and solving further gives and .
Hence the area of the circumcircle is
Find the area of rhombus given that the radii of the circumcircles of triangles and are and respectively.
The radius of the circumcircle of triangle is
The radius of the circumcircle of triangle is
Taking we get
The area of the rhombus is . Using the formula for the radius of the circumcircle of a triangle, we can write the area of as
where the side length of the rhombus can be written as . Since twice the area of this triangle is equal to the area of the rhombus, we have
This means and . Therefore, the area of the rhombus is 400.
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.
is the circumcenter of if and only if
- and when is acute and are on the same side of ;
- and when is obtuse and are on opposite sides of .
Let's see some of these criteria in action.
Quadrilateral satisfies , and . Find
Since and the angles of the quadrilateral are given, it does not hurt to see if is a circumcenter.
By the definition of a quadrilateral we know are on opposite sides of . In addition,
Then, from criterion number 3 above, we know that is the circumcenter of .
Therefore, , and our answer is
Note: As one can see, realizing that is the circumcenter of 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 be a right-angled triangle with Let be the altitude from on to Let and be the incenters of triangles and respectively. Show that the circumcenter of triangle lies on the hypotenuse .
Suppose is the circumcenter of triangle . Since , by the angle properties of cirumcenter . Note that . Hence are concyclic.
Since , externally bisects . However, is also the external bisector of . In conclusion, must lie on
The following is an advanced example that requires you to recognize the circumcenter and utilize it to perform crucial angle chasing.
is a right triangle with . is a point inside such that and is on satisfying . Also, is the point where the circumcircle of intersects
Prove that .
We want to show that . Since , it suffices to show that .
We almost want to say they are opposite angles of a cyclic quad, but lie on the same side of , so that won't be possible. This motivates the construction of as the reflection of in the line , and it suffices to show that is a cyclic quad. There are many different approaches that we could use, and we will show that . We pick this angle because is fixed in this question, while are variable points.
Now, observe that so is the circumcenter of . This means that . This motivates constructing as the midpoint of , so that we get Hence, it suffices to show that .
Observe that since they are both perpendicular to . We then have . This is equal to due to isosceles triangle . Hence we are done.
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.
Three circles with the same radius intersect at the center of a bigger circle with twice the radius.
Lines are drawn from the intersection points where two small circles meet on the big circle such that they form a triangle, as shown above left. The triangle is then divided into 3 quadrilaterals by drawing the lines from the center of the big circle to the intersection points of the small circles: red, blue, and green regions.
Alternatively, the triangle can be divided into 3 smaller triangles by simply drawing the lines from the center to the vertices: yellow, purple, and cyan regions in the diagram on the right.
Suppose that the ratio of the area of the red region, the area of the dark blue region, and the area of the green region is .
Then, if the ratio of area of the yellow region, the area of the purple region, and the area of the light blue region is , where , compute .
Let be the midpoint of side of acute triangle with and let be the circumcircle of . The tangents to at points and meet at , and and intersect at . Also, let be the altitude of and the circumcircle of . Given that and intersect at prove that
Since , we have which implies are concyclic. Since are given to be concyclic, by the alternate segment theorem . Applying the properties of cyclic quadrilaterals and alternate segment theorem again, we have
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:
Convex quadrilateral has . Point is the foot of the perpendicular from to . Points and lie on sides and , respectively, such that lies inside and
Prove that line is tangent to the circumcircle of .
First of all, points are defined in an unconventional manner, so let's try to tackle them first. Rearranging the angle equality gives
Seeing this reminds us of opposite angles in a cyclic quadrilateral summing to and is enough to motivate us to reflect over to obtain with cyclic respectively.
Now there are several ways to show that the problem is equivalent to proving , one of which will be presented at the end.
Project onto to get respectively. Right away we observe cyclic . Notice , and since is a right triangle at , bisects . Similarly we can show that bisects , implying intersection on the midpoint of , denoted by , or the circumcenter of . It is well known that the reflection of over the perpendicular bisector of lies on , and hence lies on perpendicular bisector of , which is sufficient to conclude that is an isosceles trapezoid, a cyclic quadrilateral.
Consider the radical axis of , which is perpendicular to We then know lies on it by radical axis theorem. Consequently line is the radical axis, establishing that . Hence
By the alternate segment theorem, converse is tangent to the circumcircle of and we are done.
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 be a triangle with and let be the midpoint of . The angle bisector of intersects the circle through and at the point inside the triangle . The line intersects the circle through and in two points and . The lines and meet at a point , and the lines and meet at a point . Show that is the incentre of triangle
From cyclic and symmetry, we can show which implies bisects . By the properties of incenter, is the incenter of . Note that
So now it suffices to show .
Reflect over to obtain . Now are on the same side of , so we just need to show are concyclic, which is equivalent to . Since internally bisects by symmetry, we want to prove externally bisects , or since . The last equivalence amounts to proving , which is true by
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:
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 of the incircle is called the Incenter and the radius 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.