Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Tuesday, December 6, 2016

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

Labels:
arc,
circle,
incircle,
inradius,
measurement,
radius,
right triangle,
square,
tangent

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Solution of problem 1292

ReplyDeleteWe 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;

r1/r= ME/ED=EN/EA=MN/AD= α

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

Refer to last 4 lines of your solution.

Deletehow 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

https://goo.gl/photos/jk5hvVT2HwxfZPv9A

ReplyDeleteLet 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

This is a special case of general Sangaku problem and the general solution is

ReplyDeletex= 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

Question posed from Peter Tran about problem 1292 regard last 4 lines of my solution.

ReplyDeleteI 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.

AD2= 12+ 12 AD=√2 ; area∆ AQD= (1x1)/2 =1/2 ; p=1+AD/2 = 1+√2/2 = (2+√2)/2;

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

Thank you for your explanation.

DeletePer 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 ?

Please provide more detail

See below for the sketch as per your solution

https://goo.gl/photos/8FNGp86bfDVgp2jPA