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.

## Monday, March 30, 2015

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

Labels:
circle,
circumcircle,
incircle,
inscribed,
radius,
right triangle,
tangent

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HBMO1 is a rectangle.

ReplyDeleteHD = 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.

Referring to lines " Also <BHM = <BMH = ½ HO1M = 45 deg.

DeleteFollows 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.

http://s25.postimg.org/59a5jt4qn/pro_1104.png

ReplyDeleteThis 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

Draw OL // BC where L is on O1M

ReplyDeleteApply 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