# How can every triangle have a circumcircle

Let's take for example $\triangle ABC$ with $\angle A = \angle B = 1^o$. How can a triangle like this have a circumcircle? My confusion is with triangles like this in general, with very long sides.

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How many non-collinear points do you need so that a unique circle passes through them? :) – Isomorphism Dec 18 '12 at 17:02
No matter how long you make the sides, I can make a circle big enough to circumscribe your triangle. We have all of infinity to play with, so size matters not (to paraphrase Yoda). – Todd Wilcox Dec 18 '12 at 17:04
Maybe visualise it like this: imagine any triangle with a circumcircle, with one vertex at the top. Now imagine moving the other two points round the circle - getting nearer to the top. That way you can get a triangle with 2 angles of just one degree. Then just enlarge the diagram. – Old John Dec 18 '12 at 17:06
Oh, I think I visualize it now.. so those points would be relative close together on the circle? – geometry Dec 18 '12 at 17:06
Yes, they would need to be comparatively close together. – Old John Dec 18 '12 at 17:10

Here's a link to an interactive webpage: triangle circumscribed by a circle, where you can manipulate the shape and size of the triangle, any which way you want, and each manipulation yields the unique circle that circumscribes that particular triangle.

The circumcircle always passes through all three vertices of a triangle. Its center is at the point where all the perpendicular bisectors of the triangle's sides meet (intersect). This center is called the circumcenter.

Note that the center of the circle can be inside or outside of the triangle.

Below you'll see a screen-shot from the site linked above showing the arc of a huge circle (radius 68.25) circumscribing a triangle of the sort you asked, with two sides at length 19.5 units, one side at 38.6 units.

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I think this is what the OP needed - a visualisation, rather than a proof. – Daniel Littlewood Dec 18 '12 at 19:57

Take one side of the triangle, say $AB$, and construct the perpendicular bisector $L$. Any point $P$ on $L$ will be equidistant from A and B, because $\triangle PAB$ will be isosceles by construction.

Do the same with side $BC$ and perpendicular bisector $M$.

Then the lines $L$ and $M$ are not parallel since $\triangle ABC$ is non-degenerate. Hence they intersect in a point $Q$.

$Q$ lies on $L$ so is equidistant from $A$ and $B$. $Q$ also lies on $M$ so it is equidistant from $B$ and $C$. Hence the $\bar{AQ}=\bar{BQ}=\bar{CQ}$ and $Q$ is the centre of a circle which passes through the points $A, B \text{ and } C$.

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Note that if the sides are "nearly parallel" the perpendicular bisectors will be "nearly parallel" too, but they will eventually meet. – Mark Bennet Dec 18 '12 at 17:14

For your particular triangle, let $C$ be the remaining vertex. Draw the line $\ell$ through $C$ perpendicular to $AB$. Suppose that this line meets $AB$ at $M$.

Travel on $\ell$, away from $M$, in the direction opposite to $C$. Always when $P$ is on $\ell$, $P$ is at equal distances from $A$ and $B$.

At first $P$ is much closer to $C$ than to $A$ or $B$. But if you travel to a point $P'$ that makes $\angle CAP'$ equal to $90$ degrees, $P'$ will be closer to $A$ than it is to $C$. This is because then $P'C$ is the hypotenuse of a right triangle, and $P'A$ is one of the legs of that triangle.

So $P$ somewhere between $M$ and $P'$ is "just right," it makes $PA=PB=PC$. This is the centre of our circumcircle.

Probably your discomfort is due to the fact that the centre of the circumcircle is outside the triangle. That happens whenever one of the angles of the triangle is $\gt 90^\circ$.

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It might be a little easier to understand to the problem backwards. Imagine that you constructed the circumcircle of $ABC$.

Now, $\angle A=1^\circ$ means $\arc{BC} =2^\circ$. Thus the arc $BC$ represents a 180-th of a circle. Same for the arc $AC$. Thus, the arc $ACB$ represents the 90th part of the circle, thus it is extremely tiny compared with the circle.

Moreover, starting with any circle, and picking an arc $AB$ which is the 90-th part of the circle, and then picking $C$ the midpoint of this arc, creates a triangle ABC similar to your triangle..

This explains the issue which you probably have: for such a triangle, the circumcircle has to be extremely (maybe unreasonably) big compared with the triangle...

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