## Tuesday, July 24, 2012

### Problem 790: Right Triangle, Altitude, Angle Bisector, Incircle, Tangency Point, Inradius, Circles, Tangent Line, Congruence

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

Click the figure below to see the complete problem 790.

1. http://img269.imageshack.us/img269/4894/problem790.png

Draw lines per attached sketch
We have DE perpen. to AC
angle(HBD)= angle(DBE)= angle(BDE) => EB=ED
Triangle ABE congruence to tri. AED (Right triangle Case HL)
So AB=AD => A, I1, I and E are collinear
And IB=ID= r.sqrt(2)
Right triangles AHB, HBC and ABC are similar
And r1/r=AB/AC= cos(A)
r2/r=BC/AC= sin(A)
In right triangle I1HI2 we have (I1I2)^2=(I1H)^2+I2H)^2
=2(r1)^2 +2(r2)^2 = 2.r^2 (cosA^2+sinA^2)=2.r^2
So I1I2=r.sqrt(2)

2. BI=r*sqrt(2) is obvious since angle IBA=45 degrees.
DI=r*sqrt(2) is followed from Problem 789 that "DE=r".

Let AB=a, BC=b, AC=c.

Then
BH=ab/c
AH=a^2/c
CH=b^2/c
r=ab/(a+b+c)

Denoted the radius of circle I1 and I2 by r1, r2 respectively.

Then
r1=[a^3 b/c^2] / [a + ab/c + a^2/c] = (a/c)*r
r2=[a b^3/c^2] / [b + ab/c + b^2/c] = (b/c)*r

I1H = r1*sqrt(2) = (a/c)*r*sqrt(2)
I2H = r2*sqrt(2) = (b/c)*r*sqrt(2)

Since angle I1HI2 = 90 degrees, using Pythagoras theorem,
I1I2 = r*sqrt(2)

3. BI = r*sqrt2 is straight forward by right-angled triangle and the properties of incircle.

DI = r*sqrt2 is also obvious by the previous problem 789.

For I1I2, Let AC = b, angle BCA = x, then BH = b*sinx*cosx, AH = b*(sinx)^2, CH = b*(cosx)^2

then using the relationship between inradius and triangle area of ABC, AHB and BHC, we have

r = b*sinx*cosx / (1+cosx+sinx)

r1 = (b)*(sinx)^2*cosx*sinx / (1+sinx+cosx) = r*(sinx+cosx)
r2 = (b)*(cosx)^2*cosx*sinx / (1+sinx+cosx) = r*(cosx-sinx)

Now the horizontal distance between two incenters would be
x = r1 + r2 = r*(sinx+cosx)

the vertical distance between two incenters would be
y = r2 - r1 = r*(cosx-sinx)

By pyth. theorem,
I1I2 = sqrt(x^2 + y^2)
= r*sqrt(2)

q.e.d.

4. Problem 790
Let <ABI1=I1BH=BCI2=I2CA=x, <CAI1=I1AB=<HBD=DBC=y.Then <I1AI2=x+y=45, <I1BC+<BCI2=x+2y+x=2(x+y)=2.45=90 so BI1 is perpendicular in CI2. Similar
AI1 is perpendicular in BI2. But <AI1I=x+y=45.Si BI1 intersects the CI at M and
BI2 intersects the AI1 at N,then I is orthocenter of triangleBI1I2 .Βut MI=MI1
and BM=MI2.So triagle BMI=triangleI2MI1.So DI=BI=I1I2=r√2.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE