## Monday, March 30, 2015

### Problem 1104: Right Triangle, Incircle, Circumcircle, Inscribed Circle, Radius, Tangent

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

Click the diagram below to enlarge it. 1. HBMO1 is a rectangle.
HD = EM, OD = OE imply ΔHDO and ΔMEO are congruent.
So <DHO = <EMO.
Also <BHM = <BMH = ½ HO1M = 45 deg.
Follows HBMO1 is a square.
So <DOH = <EOM = 45 deg and H, O, M are collinear.
Implies HBMO1 is a square and O is its center.
Hence x = O1M = 2 OE =2r = b + c – a.

1. Referring to lines " Also <BHM = <BMH = ½ HO1M = 45 deg.
Follows HBMO1 is a square.
So <DOH = <EOM = 45 deg and H, O, M are collinear."
Note that <DOH = <EOM = 45 deg if H, O, M are collinnear . Please provide explanation.
Note that the solution never use given data " circle O, tangent to circle Q" . Please check it.

2. http://s25.postimg.org/59a5jt4qn/pro_1104.png
This is a special case of general Sangaku problem and
the general solution is
x= r+ 2.d.(s-a).(s-c)/(b.s)…………..(1)
see sketch below for detail
http://s25.postimg.org/vvmm8s8xr/General_Sangaku_problem.png
In our case we have d= ½. b.
replace it in (1) and simplify we get
x= r+ (s-a).(s-c)/s ………(2)
in right triangle ABC (s-a)= ½(-a+b+c) and (s-c)= ½(a+b-c)
so (s-a).(s-c)= ¼.(b^2- (a-c)^2)
replace b^2= a^2+c^2 in above and simplify
we get (s-a).(s-c)= ½. ac= area of triangle ABC= s.r
so (s-a)(s-c)/s= r
and x= 2.r= a+c-b

3. Draw OL // BC where L is on O1M

Apply Pythagoras to right Tr. O1OL

(b/2 - x)^2 = (x-c/2)^2 + (a/2-x)^2 from which we have upon simplification using b^2=c^2 + a^2,

x = a+c-b which is anyway = to 2r since BDOE is a square of side r and AF = c-r and FC = a-r

Sumith Peiris
Moratuwa
Sri Lanka