geometry unbound (edited version)

8/6/2019 Geometry Unbound (edited version)

1/5

A fascinating problem in Geometry

Hoang Quoc Viet

Department of Mathematics

University of Auckland

June 20, 2011

Abstract

Geometry is an old yet interesting subject in Mathematics. For over50 years, Euclidean geometry is a compulsory topic in IMO contest (Inter-national Mathematical Olympiad). Hence, in this article, we will exploresome properties of tangently inscribed circles and wonderful facts behindit. Nevertheless, I presume that you have a bit knowledge about power-point to a circle.

1 Interesting facts

Lemma 1. Let ABC be the triangle inscribed a circle (O) where A is the com-mon intersection of another circle internally tangent to (O) and touch BC at apoint D. Then we have AD bisects the arc BC (i.e AD is the internal bisectorof the CAB

1


2/5

Proof. In order to prove this great lemma, we shall start to construct the ex-ternal tangent to both circles (denoted Ax). Let {E} = AD (O).

Hence, we have the following properties

DAx = BDA (1)

DAx = ECA (2)

Thus, from (1) and (2), it implies that BDA = ECA. Now, we continue tocompare angles

BDA = CAD +DCA (outter angle of a triangle) (3)

ECA = CEB +DCA (4)

= DAB +DCA (5)

Thus, from (3) and (5), it follows that DAB = CAD. Accordingly, AD isthe angle bisector ofCAB . We are done with the first one.

Lemma 2. Let ABC be the triangle inscribed circle (O). I is the incenter ofABC. CI meets (O) at H. (I) touches BC at D and F is the intersection ofHD and (O). Thus, we have CF IF.

Proof. Let E be the intersecting point of FI and (O) (E

=F

). Thus, if we canshow that (this statement is extremely important in the method I temporarilycall Backward Analysis) ICF+FIC = 900, we will finish the proof.

We consider the following

HA2 = HI2 = HD HF (power point H to (AFD))

2


3/5

Hence, it implies that H is a powerpoint to the cirle (DIF). Accordingly, weobtain

IFD = DIH= IHO (OH ID)

We have

FIC = EIH = IFD + IHF (6)

= FIC = IHO +IHF = OHF (7)

ICF = ICA +ACF (8)

ACF = AHF (9)

Hence, from (6) and (7), we have

FIC+ ICF = ICA+ AHF+ OHF

= BCA2

+ AHO

=BCA

2+AHB

2

=1800

2= 900

Finally, we complete our proof for the second lemma here.

2 Applications

Problem 1. Tap chi toan tuoi tho (Children Fun Maths Journal 2011 May2011)

ABC is a triangle inscribed in a circle with center ) and I is the center of theincircle of the triangle. Let G be the point of tangency of BC and (I). Constructa circle of center (Ia) touching (I) at G and internally tangent to (O) at F. AFmeets BC at D. Prove thatAID = 900.

Proof. Using the lemma 1, we easily see that FG bisects the smaller arc BC andby lemma 2, we also have IF AF .Hence, IFDG is a concyclic quadrilateral. Let J= AI (O). It greatly impliesthat

DGF = DIF

Nonetheless, we still break down angles to consider further

DGF =

BJF +

CBJ (outter angle of BJG) (10)= FAB + IAB (Since IAB = CBJ= CAB

2) (11)

= IAD (12)

3


4/5

Moreover, as mentioned above, by lemma 2,

IAD +FIA = 900

= DIF+FIA = 900

= AID = 900

Thus, our proof is complete.

Problem 2. (Unknown source)Let ABC be the inscribed triangle of a circle (O). M is a point on BC.

Construct a circle tangent to MA, MB at P and Q respectively and touches (O)at D. DQ meets (O) at E. AE intersects PQ at I. Prove that: I is the center ofthe incircle of ABC.

4


5/5

Proof. Clearly, due to the lemma 1, we have AE bisects the smaller arc BC.Thus, it suffices to show that EI = EB (why?). Since BEQ DEB , it

implies thatBE

QE=DE

BE= BE2 = QE.DE. Hence, it is enought to show

that EI2 = EQ.ED.

Claim 1.

EI2 = EQ.ED IDE= EIQ

Claim 2.ADIP is a concyclic quadrilateral

Proof. Try to show that QIE PDK. It is necessary to show KPD =EID

Using the above 2 claims, we have the desried result.

This article is dedicated to my family, without their immense encour-

agement, this article could not be published.

5

geometry unbound (edited version)

Documents