Inscribed and Circumscribed
A triangle has sides of 6, 8, and 10 inches.
What is the distance between incenter and circumcenter of the triangle?
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4 solutions
There's a simpler solution . We know that the circumcenter of a right triangle lies on the midpoint of the hypotenuse . Line joining circumcenter and hypotenuse passes 6cm from the top so we get a smaller triangle with sides as 1,r and D .
Find incenter using r(a+b+c)/2=Area
Then using Pythagoras theorem simply , D^2=(1)^2 + (r)^2
Hence D=(5)^1/2 .
Alternatively, place the vertices of the triangle at (0,0), (8,0), and (0,6). Then the centres of the inscribed and circumscribed circles are at (2,2) and (4,3), respectively, so that the distance is 5 (I used the same approach to find r = 2 )
A simpler solution
The triangle is a right triangle b/c 6^2 + 8^2 = 10^2
The hypotenuse of a right triangle is a diamter of its circumcenter so the circumcenter is the midpoint of the hypotenuse
The distances from each vertex to the incenter tangent point of each limb are equal
Let r = the incenter radius = the distance from the right angle vertex’s incenter tangents
The incenter tangent distance from the other two vertices are 6-r (top) and 8-r (right).
The hypotenuse = (6 - r) + (8 - r) = 10 -> r = 2
The coordinates of the incenter and circumcenter are (2,2) and (4,3)
The distance between these points is sqrt((4 - 2)^2 + (3 - 2)^2) = sqrt(5)
As the sides of 6 and 8 and 10 form a right triangle, the hypotenuse is the diameter of the circumscribed circle.
If we draw the radii of the inscribed circle to the sides of the triangle, we will have them perpendicular to each of the side. from there, we will have a small square at the bottom left corner of the triangle (because it has 3 right angles and 2 equal adjacent sides, which are the radii of the inscribed circle).
We call the right corner point A, the other end of the 6cm side B, other end of 8cm side C, D E F respectively the perpendicular points of 6 8 10, I the center of inscribed circle. We have:
A D + B D = 6 A E + E C = 8 B F + F C = 1 0 A D = A E ; C E = C F ; B F = B D
Hence we can say B D + C E = 1 0 .
While B D + A D + A E + C E = 1 4 ( A B + A C ) , therefore A D + A E = 4 ⇒ A D = A E = 2 ⇒ I D = I E = I F = 2 .
We call O the center of circumscribed circle. Thus B O = 5 .
And we have
B
F
=
B
D
=
A
B
−
A
D
=
6
−
2
=
4
. Thus
O
F
=
1
.
Using Pythagorean Theorem for triangle I F O , we see that O F 2 + I F 2 = I O 2 ⇒ I O 2 = 5 ⇒ I O = 5 . □
Why you are saying that CE = CF
5 , assuming the hypotenuse of the green triangle is the diameter of the blue circle.
which it is. if you force the hypotenuse to be anything other than the diameter, 1 vertex of the triangle would not be on the circle
How can we know that the distance is 5 just by knowing that the hypotenuse of the green triangle is the diameter of the blue circle?
Applying the formula for the inradius of a triangle ( s a r e a ) gives us inradius = 2 ⋅ 1 2 6 ⋅ 8 = 2 . Also, we know that the circumcentre of a right triangle is the midpoint of the hypotenuse. Let this be point A .
The length of the tangent in red is s − 8 = 4 . ⇒ Length of the green line segment is 5 − 4 = 1 .
So, distance between circumcenter and incenter = A B = 5
The inradius of any triangle is given by r = a + b + c 2 [ A B C ] , and the circumradius by R = 4 [ A B C ] a b c .
In this case, the triangle is right because 6 2 + 8 2 = 1 0 2 , so the area is simply [ A B C ] = 2 6 × 8 = 2 4 .
So, r = 6 + 8 + 1 0 2 × 2 4 = 2 and R = 4 × 2 4 6 × 8 × 1 0 = 5 .
Finally, we apply Euler's theorem in geometry, where d is the distance we are looking for:
d = R ( R − 2 r ) = 5 ( 5 − 2 × 2 ) = 5 .