## Monday, May 19, 2008

### Elearn Geometry Problem 41

See complete Problem 41
Sangaku, Mickey Mouse, Circles, Tangent. Level: High School, SAT Prep, College geometry

1. Using Pythagoras, BC^2 = d^2 + (c - b)^2
Let angle CAB = z
Using cosine rule: BC^2 = (a+c)^2 + (a+b)^2 - 2(a+c)(a+b)cos(z)
Or d^2 + (c - b)^2 = (a+c)^2 + ((a+b)^2 - 2(a+c)(a+b)cos(z)
From where, d^2+ b^2+c^2 - 2bc = a^2+c^2+2ac + a^2+b^2+2ab - 2(a+c)(a+b) cos(z)
Or d^2=2a^2+2ac+2ab+2ac-2(a+c)(a+b)cos(z)
Or d^2=2(a+b)(a+c)-2(a+c)(a+b)cos(z) or d^2
= 2(a+b)(a+c)(1-cos(z) --------(i)
From the triangle in circle A we can write,
x^2 =a^2 +a^2-2a^2*cos(z)
=2a^2(1-cos(z)
Or(1-cos(z)) = x^2/2a^2-------------(ii)
From (i) and (ii), we get:
d^2 = 2(a+b)(a+c)x^2/2a^2
= (a+b)(a+c)x^2/a^2
Or x^2 = a^2*d^2/[(a+b)(a+c)]
QED
Ajit:ajitathle@gmail.com

2. bae deok rak(bdr@korea.com)October 7, 2010 at 10:05 PM

By cosine rule,
d^2 +(b-c)^2 =BC^2
=(a+b)^2+(a+c)^2-2(a+b)(a+c)cos(ang(BAC))
=(a+b)^2+(a+c)^2-2(a+b)(a+c)(a^2+a^2-x^2)/2aa
=2a^2+b^2+c^2+2ab+2ac-2(a^2+ab+ac+bc)
+(a+b)(a+c)(x^2/a^2)
<=>
d^2=(a+b)(a+c)(x^2/a^2).

We get, x^2 =(a^2 d^2)/(a+b)(a+c)

3. From Moscow: I have pure synthetic solution but not sure if correct. Please check. Dropping perpendiculars from A, F, and G onto DE to get A' F' G', we have 1)A'F'/A'D=a/(a+b)and 2)A'G'/A'E=a/(a+c). Looking at pentagon ABCDE, We note that angle A + angle B + angle C + 90 + 90= 540, so angles A, B, and C add to 360. We see that angle FDE is half of B and angle GED is half of C. Therefore FD and GE meet at point P such that angle FPG is half of angle A so P is on circle A. But angle BDF = 90-angle FDE = BFD = AFP = APF, so AP is perpendicular to DE. x/d=PF/PE=PG/PD, so x^2/d^2=(PF*PG)/(PE*PD)=(A'F'*A'G')/(A'D*A'E)=a^2/((a+b)(a+c)). The last two equality statements we get from 1) and 2) and we are done.

1. To Ivan from Moscow, Problem 41. Your great pure synthetic solution is correct. To complete your solution, what is the reason for the step: x/d=PF/PE=PG/PD. Thanks.