how to find the distance of the incenter of an equlateral triangle to mid center of each side?

Question

I am given a diagram of a equalteral triangle with all bisectors drawn, the incenter and the circumcenter are one and the same. I am also given the distance of the circum center from the vertexes. I am told to find the distance from the incenter to the midpoint of each side, how do I solve it?

Answer 1

The best thing to do is to draw this triangle on your paper first. If you draw each of the angle bisectors to the incenter, you will create 6 right triangles. Since the angle bisectors create a 30 degree angle with the side of the triangle, and the angle bisector creates a 90 degree angle where it bisects the side of the triangle, you have a 30-60-90 triangle. If the distance from the incenter to the vertex is given... lets let that = c. This is the hypontenuse of the right triangles you created in your drawing. The distance to the midpoint is the smallest side of your right triangle, so this would be .5c using the properties of a 30-60-90 triangle.

Answer 2

The distance of the circumcenter to the vertex is two thirds of the altitude. The length you are trying to find is the other third. Just halve the distance you already know. 

Answer 3

All four of the principal concurrence points of a triangle coincide in an equilateral triangle. The ratio of base to centroid compared to centroid to vertex distance is 1 to 2.

So the ratio of the incenter to base compared with circumcenter to vertex distance is also 1 to 2 in an equilateral triangle. This is the same conclusion as many of the other posters have reached.

The use of 30-60-90 right triangles is also valid.

In other words, the radius of the circle around the triangle has a radius twice as large as the circle inside the triangle.

PS: Equiangular triangles are another term for equilateral triangles. Just less commonly used.

PPS: The ratio of 1 to 2 for the distance from base point to centroid compared to the distance from centroid to vertex is only valid IN GENERAL for the centroid. This ratio can be used for equilateral triangles in this case because all four major points of triangular concurrency (incenter, circumcenter, centroid, and orthocenter) coincide.

Answer 4

We have an equilateral triangle with side a, so the The perpendicular bisector of side (which is also the median, the altitude and the angle bisector of the triangle) is a radical 3 /2 and the distance from the center of a side to the center of the triangle is just 1/3 of it. (This is the property of the incenter point of the triangle which is the intersection of the medians). So:

a Radical 3 /2 x 1/3 = a Radical 3 / 6.

Answer 5

An equilateral triangle has 3 equal sides and 3 equal angles. Since the sum total of any triangle is 180 degrees, then each angle is 1/3 x 180 = 60 degrees. With the triangle having one side as the bottom or the base, drop a line vertically down from the triangle center (point C) to the bottom at the point which bisects that bottom side (point B). Draw another line from the center (C) to the angle at the left, (angle or point A). ABC is now a triangle with angle A = 30 degrees, angle B = 90 degrees, and Angle C as 60 degrees. The sides of this smaller triangle ABC are in a ratio of AC/CB = 2/1 AB/CB = 3^1/2/1 If the hypotenuse = 2, then CB = 1 Let a side of the equilateral triangle = x. Then AB=x/2. The distance from the center of the equilateral triangle to the bisection of a side = CB. AB/CB = (x/2)/CB = 3^1/2/1 = 3^1/2 Solve for CB = (x/2)/3^1/2 = x/2(3)^1/2. CB = half of the equilateral triangle's side divided by the square root of 3. For example, if a side of the equilateral triangle is 2, then the distance from its center to the bisection of the bottom is 1/2 times 2 = 1 divided by the square root of 3. CB would then be 1/3^1/2.

Answer 6

If the triangle is equilateral, then you form a 30, 60, 90 triangle. Use the bisector to the vertex as a hypotenuse with length equal to two times the length of the shortest side. The shortest sideis the distance from the circumcenter to the the vertexes. All you have to do is take the distance given to you and divide it by two.

Answer 7

The first thing you have to keep in mind is the symmetry of the problem. Here you have an equiliateral triangle, meaning all sides have the same length and all angles measure the same. That gives you the first clue, which is that the angles of the triangle are all 60 degrees (360/3).

The second thing you have to realize is that any line from a vertex to the circumcenter will bisect the angle of 60 degrees into two 30 degree angles. Put this together with the fact that a line that goes through the midpoint of one of the sides of the triangle and the circumcenter.

If you draw all this out you will find out that you wind up with a nice set up to the problem, where you are looking for the length of one of the sides of a right triangle.

The final step is to use trigonometry. You'll have a right triangle with an angle of 30 degrees, and a hypotenuse equal in length to the distance between the circumcenter and the vertex. Solve for the side they are asking about (think about the sine and cosine definitions), and you should be set.