Geometry Problem

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

Click the figure below to see the complete problem 792.

## Thursday, July 26, 2012

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

Labels:
altitude,
angle bisector,
concyclic,
incenter,
incircle,
right triangle

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

ReplyDeleteDraw 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

First of all, since

ReplyDeleteangle ABD

= angle ABH + angle HBD

= angle BCH + angle CBD

= anle ADB

So AB = AD.

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.

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

ReplyDeleteBy 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.

In addition,

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.

∠EBD=90°:2=45°

ReplyDeleteΔ 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

[ Erina from NJ]

ReplyDeleteAngle 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

Angle EBD=90°:2=45°

ReplyDeleteΔ 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....

Problem 792

ReplyDeleteIn 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