Thursday, July 26, 2012

Problem 792: Right Triangle, Altitude, Angle Bisector, Three Incircles, Five Concyclic Points, Circles

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

Click the figure below to see the complete problem 792.

1. http://img88.imageshack.us/img88/5497/problem792.png

Draw lines per attached sketch
Per the result of problem 789 we have FD=FI= FE= inradius of triangle ABC
In Problem 789 we prove that triangle BEC is isosceles , C, I ,I2 are collinear and CI is the perpendicular bisector of BE .
We have ∠LBI2=45
Due to symmetric property of symmetric line CIL, we have ∠ LEI2=∠LBI2=45 => EI2 perpendicular to BD
Similarly we also have DI1 perpendicular to BE
So E , I1,I, I2 and D are concyclic in a circle diameter ED

2. First of all, since
angle ABD
= angle ABH + angle HBD
= angle BCH + angle CBD

Now since AI1 is angle bisector of angle BAD,
and triangle ABD is isosceles,
thus AI1 is perpendicular to BD.

Consider triangle I1BD,
from the above result, it is isosceles.
Since I1B = I1D, and angle I1BD = 45,
so angle BI1D = 90.

Therefore, angle EI1D = 90.
Similarly, angle EI2D = 90.

(Indeed, BH, DI1 and EI2 are concurrent,
and they intersect at the orthocenter of triangle BED. )

On the other hand,
since AI1 is perpendicular to BD,
AI is also perpendicular to BD, so IB = ID.

Similarly, IB = IE. So IB = ID = IE.
Therefore, I is the circumcenter of triangle BED.

Consequently,
angle EID = 2 * angle EBD = 90.

As angle EI1D = angle EID = angle EI2D = 90,
thus E, I1, I, I2, D concyclic, with diameter DE.

3. Simply let AB = a, BC = b and AC = c.

By Van Aubel theorem, on triangle ABH and BHC,
BI1/I1E = (BH+BA)/AH = (b+c)/a,
BI2/I2D = (BH+BC)/HC = (a+c)/b,

By angle bisector theorem,
EH = BH/(BA+BH)*AH = (ab/c)*(a/(b+c))
HD = BH/(BH+BC)*HC = (ab/c)*(b/(a+c))

Hence EH/HD = (a(a+c))/(b(b+c))
And further,
(BI1/I1E) * (BI2/I2D) * (EH/HD) = 1

So BH, I1D, I2E are concurrent.

S(BOD)/S(DOE) = BI1/I1E = BH/EH = tan(BED)
S(BOE)/S(DOE) = BI2/I2D = BH/DH = tan(BDE)

As a result, by the barycentric coordinate system on triangle BDE, the concurrent point is the orthocenter.
So Angle EI1D = Angle DI2E = 90 and I1, I2, D, E are concyclic.

For the remaining, EBD = 45, while BI = ID = IE , so I is the circumcenter of BDE, and IDE = 90.

So the point I also lies on the circle I1, I2, D, E.

Q.E.D.

4. ∠EBD=90°:2=45°
Δ ABD, Δ CEB=> isosceles
AI₁ I =>perpendicular bisector [BD] =>I₁D ┴ BE
­ CI2I → perpendicular bisector [BE] =>EI2 ┴ BD
IM ┴ AC MD=r EM=r IM=r

Points E, I₁, I, I2 and D are in the circles with the center M and radius r

5. [ Erina from NJ]

Angle EBD=90°:2=45°
Δ ABD,Δ CEB => isosceles
AI₁ I => perpendicular bisector [BD]=> I₁D ┴ BE
­CI2I → perpendicular bisector [BE]=> EI2 ┴ BD
IM ┴ AC MD=r EM=r IM=r

Points E, I₁, I, I2 and D are in the circles with the center M and radius r

6. Angle EBD=90°:2=45°
Δ ABD,Δ CEB => isosceles
AI₁ I =>perpendicular bisector [BD] =>I₁D ┴ BE
­ CI2I → perpendicular bisector [BE] =>EIₐ ┴ BD
IM ┴ AC MD=r EM=r IM=r

Points E, I₁, I, I2 and D are in the circles with the center M and radius r

...ERINA....

7. Problem 792
In the same figure with the problem we have 790. <BI1I=45=<IDE then the points E,I1,I,D are
Cyclic.But B,E are symmetrical CF so <IEI2=IBI2=<IDI2.So the points E,I,I2,D are cyclic.
Therefore E,I1,I,I2 and D are concyclic.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE