## Tuesday, December 6, 2016

### Geometry Problem 1292 Square, Circle, Right Triangle, Arc, Inradius, Radius, Incircle, Measurement

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

1. Solution of problem 1292
We call AE=b; ED=c; AD=a; F is the perpendicular to AD from Q that meet circle ABCD
K is the point of tangent of circle O1 parallel to AD that meet extension of ED and EA at M and N
H is the point of cross AD and QF; AH is the radius of circle AQD; G is the point that cross MN with QF;
radius r = area ∆ AED/p in which p= (a+b+c)/2
radius r1= area ∆ EMN/p' in which p’= (a+b+c) α/2 and so
r1/r= α for all the position in which E is on the circle AQD.
We need now to calculate α.
If we consider the right triangle AQD and put AQ=QD=QF= 1 = radius of circle ABCD;
we have AD= √2; r= 1/(2+√2)
M’ and N’ the point in which the tangent to F cross the extension of QD and QA
FM’=FN’=1; M’N’=2; QM’=QN’= √2
r1= 1/(1+√2) ; r1/r= (2+√2)/(1+√2) ; (r1)2/(r)2= (2+4+4√2)/(1+2+2√2)=2 and so r1/r= α =√2

1. Refer to last 4 lines of your solution.
how do you get r= 1/(2+√2) and r1= 1/(1+√2) ?
Is r and r1 depend on value of AE and ED ? Please explain

Peter

2. https://goo.gl/photos/jk5hvVT2HwxfZPv9A

Let e=AD, d=AE, a=ED and x= r1
Observe that AQED is cyclic and AD and QN are diameters
Q, O, M are collinear and EO meet circumcircle of AED at N
∠ (QEN)= 90 degrees
And QA=QD=AN=ND=e /sqrt(2)
We have r=1/2(a+d-e)
Apply Ptolemy’s theorem in qua. AQED we have
QE.AD + AQ.ED=AE.QD => QE= (d-a)/sqrt(2)
Apply Pythagoras’s theorem in tri. QO1E we have
(e/sqrt(2)-x)^2=(x.sqrt(2))^2+(d-a)^2/2
Replace d^2+a^2=e^2 in above and simplifying we have
.x^2+sqrt(2).e.x –a.d= 0
Determinant delta= 2(a+d)^2 after replace e^2=a^2+d^2
And positive solution x= sqrt(2).(a+d-e)/2 = sqrt(2).r

3. This is a special case of general Sangaku problem and the general solution is
x= r+ 2.delta..(s-d).(s-a)/(e.s)…………..(1)
see sketch below for detail
http://s25.postimg.org/vvmm8s8xr/General_Sangaku_problem.png
In our case we have 2.delta= e.(sqrt(2)-1)
replace it in (1) and simplify we get
x= r+ (sqrt(2)-1) .(s-a).(s-d)/s ………(2)
in right triangle AED (s-a)= ½(-d+e-a) and (s-d)= ½(a+e-d)
so (s-a).(s-d)= ¼.(e^2- (d-a)^2)
replace e^2= a^2+d^2 in above and simplify
we get (s-a).(s-d)= ½. ad= area of triangle ABC= s.r
so (s-a)(s-d)/s= r
and x= r+(sqrt(2)-1).r= sqrt(2) r

4. Question posed from Peter Tran about problem 1292 regard last 4 lines of my solution.
I suppose to insert circle with radius r1 a r in triangles QN’M’ and QAD the base of which is F and H on N’M’ and AD.
r= area∆AQD/p= 1/2:(2+√2)/2 = 1/(2+√2)
N’F=M’F=QF=1 ; QM’=QN’=√2 ; area ∆N’M’Q= (√2x√2)/2 =1 ; p’ of ∆QM’N’=(√2+√2+2)/2=√2+1
r1= area∆N’M’Q/p’ = 1/(√2+1);
r1/r = 1/(√2+1/:1/(√2+2) = √2 = α
This rapport is independent of position of E on circle AQD end is independent of the dimensions of r1 and r

1. Thank you for your explanation.
Per your response, I understand that in your solution, you suppose that
. r= incircle radius of triangle QAD not triangle AED per problem statement
. r1= incircle radius of triangle QN’M’ not as per problem statement
This special case only happen when E coincide to Q.. in this case triangle AED become AQD and r1= √2. r
What happen for general case when E do not coincide to Q ? how about ratio of r1/r in this case ?