## Saturday, May 15, 2010

Geometry Problem
Click the figure below to see the complete problem 454 about Triangle, Incircle, Inradius, Tangent, Circle, Radius.

Complete Problem 454

Level: High School, SAT Prep, College geometry

1. It's easy to see that r= ra + (r+ra)sin(A/2) or ra=r(1-sin(A/2))/(1+sin(A/2)). Similarly,rb=r(1-sin(B/2))/(1+sin(B/2)) and rc=r(1-sin(C/2))/(1+sin(C/2)). Now R.H.S. = r[(1-sin(A/2)(1-sin(B/2))/(1+sin(A/2)(1+sin(B/2))]^(1/2)+(1-sin(B/2)(1-sin(C/2))/(1+sin(B/2)(1+sin(C/2))]^(1/2)+(1-sin(C/2)(1-sin(A/2))/(1+sin(C/2)(1+sin(A/2))]^(1/2)]. The expression inside the big brackets is always = 1 for A+B+C=180 as may be verified numerically(but I don't know how to prove the identity!). Thus,RHS = r. In other words,r=(ra*rb)^(1/2)+(rb*rc)^(1/2)+(rc*ra)^(1/2)
Ajit

2. Proof of the trigonometric identity.
Consider the term on RHS. Under the square root sign we've [(1-sin(A/2)(1-sin(B/2))/(1+sin(A/2)(1+sin(B/2))]
=[(1-(sin(A/2))^2)(1-(sin(B/2))^2)]/[(1+sin(A/2)(1+sin(B/2))]^2
= [cos(A/2)cos(B/2)/[(1+sin(A/2))(1+sin(A/2))]^2
Hence RHS = cos(A/2)cos(B/2)/(1+sin(A/2))(1+sin(B/2)).
Now we use the identity cos(t)/(1-sin(t))=(1+sin(t))/cos(t).
So RHS = [(1-sin(A/2))(1-sin(B/2))cos(C/2)+(1-sin(B/2))(1-sin(C/2))cos(A/2)+(1-sin(C/2))(1-sin(A/2))cos(B/2)]/ [cos(A/2)cos(B/2)cos(C/2)]
Now for the sake of ease of writing, let's call sin(A/2)=s1, sin(B/2)=s2 & sin(C/2)=s3 and cos(A/2)=c1, cos(B/2)=c2 & cos(C/2)=c3. Now on expansion RHS = [c1+c2+c3+c1s2s3+c2s1s3+c3s1s2-c1s2-c1s3-c2s1-c3s1-c3s2-c2s3/(c1c2c3). Because A/2+B/2 =90-C/2, we've c1=s2c3+s3c2, c2=s1c3+s3c1 & c3=s1c2+s2c1. On expansion and simplification, we now have RHS = (c1s2s3+c2s1s3+c3s1s2)/c1c2c3. Further recall that c1c2-s1s2=s3 once again because of A/2+B/2=90-C/2. Now RHS = [s3(c1s2+c2s1)+c3s1s2)]/c1c2c3 = [s3c3 + c3s1s2]/(c1c2c3)=[(c1c2-s1s2)c3+s1s2c3]/c1c2c3=(c1c2c3)/(c1c2c3)= 1
Ajit

3. Joe said...
It's easy to see that r= ra + (r+ra)sin(A/2) or ra=r(1-sin(A/2))/(1+sin(A/2)). Similarly,rb=r(1-sin(B/2))/(1+sin(B/2)) and rc=r(1-sin(C/2))/(1+sin(C/2)). Now R.H.S. = r[(1-sin(A/2)(1-sin(B/2))/(1+sin(A/2)(1+sin(B/2))]^(1/2)+(1-sin(B/2)(1-sin(C/2))/(1+sin(B/2)(1+sin(C/2))]^(1/2)+(1-sin(C/2)(1-sin(A/2))/(1+sin(C/2)(1+sin(A/2))]^(1/2)].

See this problem
http://demonstrations.wolfram.com/InscribingFourCirclesInATriangle/ - similar problem.

4. In the Wolfram demo. the tangents are parallel to the sides leading to the result, r=ra+rb+rc.
In the case in hand, we must prove generally that r=(rarb)^(1/2)+(rbrc)^(1/2)+(rcra)^(1/2)
How would this demo. help us?

5. Wolfram demo is identical this problem 454 for a=b=c only.

And what do You think function r(ra,rb,rc,rd)=? for circumscribed quadrilateral ABCD about?

6. For Wolfram demo we can find
sa/s and can find
ra/r=(-a+b+c)/(a+b+c)
rb/r=(+a-b+c)/(a+b+c)
rc/r=(+a+b-c)/(a+b+c).
And we can see r=ra+rb+rc.

1)What do You think function r(ra,rb,rc,rd)=? for circumscribed plane quadrilateral ABCD about?
2)what do You think function r(ra,rb,rc,rd)=? for circumscribed 3D quadrilateral ABCD - tetrahedron about?

7. We also have ra/r=(1-sin(A/2))/(1+sin(A/2))=(1-cos(90-A/2))/(1+cos(90-A/2))=2sin^2(45-A/4)/2cos^2(45-A/4)=tan^2(45-A/4). Similarly, we have rb/r=tan^2(45-B/4) and rc/r=tan^2(45-C/4). Lets call x=45-A/4, y=45-B/4, z=45-C/4. Then x+y+z=90, therefore (tan x + tan y)/(1 - tan x tan y)=tan(x+y)=tan(90-z)=1/tan z, hence tan x tan y + tan y tan z + tan z tan x = 1, therefore sqrt(ra/r*rb/r)+sqrt(rb/r*rc/r)+sqrt(rc/r*ra/r)=1, then r=sqrt(ra*rb)+sqrt(rb*rc)+sqrt(rc*ra).