Right triangles and tangent circles

[Parcival]

Inspired by Grayfalcon's post.

 

Consider a right triangle ABC, with a = 3, b = 4, and c = 5. Circles centered at A, B, and C of radii 3, 2, and 1, respectively, are pairwise externally tangent. Such circles have, to the best of my knowledge, no name. I shall refer to them as the triangle's G-circles.

 

What are the radii of the G-circles of a right triangle of sides 5, 12, and 13?

Of sides 20, 21, and 29?

[psi20]

So you have 3 circles with radii x, y, z

 

x + y = 5

y + z = 12

x + z = 13

 

When you solve it, you get x = 3, y = 2, z = 10.

 

x + y = 20

y + z = 21

x + z = 29

 

x = 14, y = 6, z = 15

[Parcival]

Right.

 

Now, moving from algebra (back) to geometry, the (not necessarily right) triangle ABC has 3 G-circles, pairwise tangent at the points D, E, and F. Let d, e, and f be the lines tangent to the pairs of G-circles at the points D, E, and F, respectively. Do d, e, and f have a point in common?

[psi20]

It took me 3 hours and 3 sheets of paper to figure it out, so I can't put all the stuff down except for the important stuff. There was also my lack of foresight in realizing that using x and y as G-circle radii wasn't a good idea. Diagrams will help in reading this at any rate.

There were some interesting results. Skip to the bottom after reading next 3 paragraphs for the the intersection of tangents.

 

My method of problem solving was to put an obtuse triangle on the coordinate plane with an endpoint called C at (0,0). Point A of triangle is in Quadrant II, and point B is on the x-axis to the right of the origin.

 

The G-circle of point C has a radius of z. The G-circle of A has radius of x. The G circle of B has radius of y.

 

The sides of the triangle opposite of the points are lower cases of the point. Meaning side "a" is on the x-axis, which is opposite of point "A" in Quadrant II. The angles are labeled by the upper case. Angle A is at point A.

 

The line containing side "a" is y=0. That means the a pairwise tangent is at x=z, perpendicular to "a" and containing the point of intersection of the 2 G-circles. The tangent will be called GTAN1

 

Define the slope of line "b" as -n/m. That means the pairwise tangent of these 2 G-circles will have a slope of m/n. Secondly, the pairwise tangent goes through the intersection of the two G-circles, which lies on the point (z cos C, z sin C) using trigonometry. Having the slope and the point, you figure that the pairwise tangent is y= (mx + zn sin C - zm cos C)/ n . That's GTAN2

 

Line "c" contains (a, 0). The x-coordinate is the distance from the origin, which is the side "a." Say that (a,0) was the center of the G-circle centered at point B. The intersection of the two circles would be ( a + y cos (A+C), y sin (A+C)). This is because angles A + B + C = 180. Therefore, 180 - B = A + C . Adding the "a" in the x-coordinate is because you shift the graph "a" to the right from the origin. From the 2 points (a,0) and (a + y cos (A+C), y sin (A+C)), we figure the slope of line "c" to be tan(A+C).

That means the slope of the pairwise tangent is -cot (A+C) going through (a + y cos (A+C), y sin (A+C)). We figure the pairwise tangent line to be

y = (-x cos(A+C) + a cos (A+C) + *y)/ sin (A+C) which is called GTAN3. Notice that I put *y. That's y the distance, or the radius of the G-circle centered at B. Bad judgement on my part.

 

Now we have 3 GTAN lines. The intersection of GTAN1 and GTAN2 is

(z, (mz + zn sin C - mz cos C)/ n) . The intersection of GTAN1 and GTAN3 is

(z, (-z cos(A+C) + a cos (A+C) + *y)/ sin (A+C))

 

Oh, using the equation y-a=-z and substituting it into the y-coordinate of the point, we get

(*y(1+ cos(A+C))/ sin (A+C))

 

If the three lines have a point in common, (mz + zn sin C - mz cos C)/ n = (*y(1+ cos(A+C))/ sin (A+C))

 

When you have acute angles, the m turns into -m, meaning the distance changes direction from the left of the origin to the right of the origin. When you have a right tirangle, m = 0. When you have an obtuse triangle, m is positive. Make sure you put the right numbers into the calculator.

 

It turns out equilateral triangles are do intersect. The obtuse triangle and 3-4-5 triangle I tried have really close intersections, but not exactly.

 

*Hold that thought, doing it manually might give more accurate results than calculators.

[psi20]

Yeah, it turns out that I subconsciously and automatically estimated the results and plugged the estimations into the calculator. Although I estimated to the thousandths, the extra decimals caused the difference at the end to appear.

 

It seems that the answer to your question is yes, the pairwise tangent lines have a point in common whether they are obtuse, right, or acute.

 

*Err, now we need a general proof rather than guess and check.

[psi20]

For right triangles, I simplified some equations down to

 

y= (ab)/(-a+b+c)

 

If someone can prove this to be true for right triangles, then the proof for right triangles will be done. It seems true for the values I put in.

[psi20]

For all triangles I've gotten it down to y = 2(area)/(perimeter - 2a)

[Parcival]

For all triangles I've gotten it down to y = 2(area)/(perimeter - 2a)

 

Not for all, surely. For an equilateral triangle with sides of length 2, y=1, but 2*area/(perimeter-2a) = 2*sqrt(3).

[Parcival]

A hint vis a vis finding y. Unfortunately, I don't see how knowing x, y, and z will help with the question as to whether the three tangents intersect.

 

[hide]perimeter/2 = x + y +z.[/hide]

[psi20]

Oh yeah, you're right . However, I'm still sure the one for right triangles is accurate. Proving that y= (ab)/(-a+b+c) will be able to show that the tangents intersect for right triangles.

 

(mz + zn sin C - mz cos C)/ n = (y(1+ cos(A+C))/ sin (A+C))

m=0, C is 90 degrees

The equation becomes

(0z + zn*1 - 0zcos90)/n = y(1+ cos(A+90))/ sin(A+90)

zn/n = y(1+ -sinA)/ cosA

 

z=y(1-sinA)/cosA

Remember that z+y=a, so z= a-y. Remember that sinA is a/c and cosA is b/c from trig.

 

Therefore (a-y) = y(1-(a/c))/(b/c)

(a-y)(b/c) = y((c-a)/c)

(ab-yb)/c = (yc-ya)/c

ab - yb = yc - ya

ab = yc - ya + yb

ab = y(-a + b + c)

y = ab/(-a + b + c)

 

The reason, bad judgement on my part, to say for all triangles that

y = 2(area)/(perimeter - 2a)

is because it works for all right triangles and an obtuse triangle I tried (*coulda sworn I did it for the equilateral, perhaps bad memory).

 

Anyways, if we can prove for right triangles that

y = ab/(-a + b + c), it means that for right triangles, (mz + zn sin C - mz cos C)/ n = (y(1+ cos(A+C))/ sin (A+C)). Working our way back up, that means the y-coordinates are all the same when the x-coordinates are the same, meaning they intersect.

[psi20]

Ok I think I have a proof for tangents of G-circles for equilateral triangles intersecting.

 

Equilateral triangle means z and y are equal, A and C are 60, tan 60 = n/m. More importantly, using my method, acute angles have negative m's.

(mz + zn sin C - mz cos C)/ n = (y(1+ cos(A+C))/ sin (A+C)) becomes

(-my+ yn*sin60 + my cos 60)/n = y(1 + cos(120))/(sin (120))

 

...basically going down the steps with algebra, cancel out the y's, do some more algebra, substitute n=m tan60,

you prove that this is true.

Therefore, tangents intersect.