Simple Problem and Its Applications

8/7/2019 Simple Problem and Its Applications

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The simple geometry problem and its applications

The problem below is well-known but its applications are very interesting and profound.

Problem 1:Let the incircle (I) of triangle ABC is tangent to AB, AC at M,N respectively. The line BI

intersects MN at E. Prove that 90o

BEC ∠ =

Solution:

We only prove this problem when E lies outside the segment MN (other case is similar)

Since (I) toughs AB, AC at M, N respectively we get AMN ANM ∠ = ∠

Then 0902 A ENC ANM ∠∠ = ∠ = −

On the other hand, we have 902 2 2

o B C A EIC IBC ICB

∠ ∠ ∠∠ = ∠ + ∠ = + = −

Therefore ENC EIC ∠ = ∠ Hence quadrilateral INEC is cyclic.

We obtain 90o

BEC IEC INC ∠ = ∠ = ∠ = (q.e.d)

Now, let’s start with some applications of problem 1 :

Problem 2:

Let the incircle (I) of triangle ABC tough BC, CA, AB at M,N,P respectively. The line BI

intersects MN,PN at G,H, CI intersects PM,PN at J,K. Show that 4 points G,H,J,K lie onone circle.

Solution:



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Using problem 1 we get 90

o

AJC AGB BKC BHC ∠ = ∠ = ∠ = ∠ =

AJIG is a cyclic quadrilateral so KJG IJG IAG∠ = ∠ = ∠ (1)

Moreover 90o

AGI ANI ∠ = ∠ = then quadrilateral AIGN is cyclic.

We get IAG ING ICN ICB KCB∠ = ∠ = ∠ = ∠ = ∠ (2)

Since 90o

BKC BHC ∠ = ∠ = , we have quadrilateral BKHC is cyclic.

Hence KCB KHB KHG∠ = ∠ = ∠ (3)

From (1),(2) and (3) we obtain KJG KHG∠ = ∠

Therefore GHJK is a cyclic quadrilateral (q.e.d)

Problem 3:

Let xBy α ∠ = . A is a fixed which lie point on the ray Bx. C moves on the ray By. The

incircle (I) of triangle ABC touches AC, BC at E,F respectively. Prove that the line EFalways passes through a fixed point.

Solution:

Let M be the intersection of EF and BI.

From problem 1, we have 090 AMB∠ =



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Note that2

ABM α

∠ = and A, B are two fixed points.

Then M is a fixed point.

Therefore the line EF passes through a fixed point M (q.e.d)

Next, we will use the problem 3 to solve this hard and nice problem: Problem 4:

Given a triangle ABC. A point D moves on the opposite ray of the ray BC such that theincircles of triangles ABD and ACD intersect each other at P and Q. Prove that the line

PQ always passes through a fixed point.

Solution:

Let (O1) tangent to BD, AD at M,N; (O2) tangent to BD,AD at X,Y respectively. BO1

intersects MN at E, CO2 intersects XY at F.

Note that the angles ABC, ACD are invariable then by problem 3, we get E,F are two

fixed points.

Because MN, PQ, XY are perpendicular to O1O2, we obtain XY//PQ//MN (1)

Let I, K be the intersections of PQ and BD,EF respectively.

We have IM2=IQ.IP=IX

2then IM=IX (2)

From (1) and (2) we get PQ is a median of trapezoid MEFX

So KE=KF=1/2EF

Since E,F are two fixed points we have K is a fixed point.



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Therefore PQ passes through a fixed point K.

Problem 5 (Vietnam NMO 2009):

Let A, B be two fixed points and C is a variable point on the plane such that ACB α ∠ =

(constant) (

0

0 180

oα < <

). Let N, M, P be the projections of the incenter I of triangleABC to its sides BC, CA, AB respectively. Denoted by E, F the intersections of AI, BI

with MN, respectively. Prove that the length of the segment EF is constant and thecircumcircle of triangle EFP always passes through a fixed point.

Solution:

a, Applying problem 1, we have 90o

AFB AEB∠ = ∠ = then the quadrilateral AFEB iscyclic.

On the other hand, 90o

INB IEB∠ = = so the quadrilateral INEB is cyclic.

We get2

ACBFBE IBE MNI

∠∠ = ∠ = ∠ =

The diameter AB of the circumcircle of quadrilateral AFEB is constant therefore EF is

constant (q.e.d)

b, It’s easy to see that the circumcircle of triangle EFP is the Euler’s circle of triangle

AIB hence it passes through the midpoint J of segment AB and J is a fixed point.

Problem 6 ( Iberoamerican 1989):



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Let the incircle (O) of triangle ABC touch AC, BC at M, N, respectively. The bisectors of

angles A and B meet MN at P, Q, respectively. Prove that MP.OA=BC.OQ

Solution:

By problem 1, we have 90o AQB APB∠ = ∠ =

Since 090 AMO AQO∠ = ∠ = we get quadrilateral AOQM is cyclic.

Then AOQ QMC MOC ∠ = ∠ = ∠

So cos cos AOQ MOC ∠ = ∠

OrOQ OM

OA OC = (1)

We need to prove thatOM MP

OC BC =

We have OMP OCB∠ = ∠

BOPN is a cyclic quadrilateral hence OPM OBC ∠ = ∠

Therefore POM BOC △ ∼△ we obtainOM MP

OC BC = (2)

From (1) and (2) we getOQ MP

OA BC =

So MP.OA=OQ.BC (q.e.d)



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Problem 7 (Romanian IMO Team Selection Test 2007):

Let ABC be a triangle, let E,F be the tangency points of the incircle (I) to the sides AC,

AB, respectively, and let M be the midpoint of the side BC. AM meets EF at N. (M, MB)

intersects BI, CI again at X,Y, respectively. Show that NX AC

NY AB

=

Solution:

Using problem 1 we have X,Y lie on EF.

Draw a line d which pass through A and parallel to BC. EF meets d at K. Let D be the

projection of I to BC. ID intersects EF and d at N’,H, respectively.

Three points F,H,E are visible from AI under the right angle so the quadrilateral HFIE is

cyclic.

Then IHF FEI EFI IHE ∠ = ∠ = ∠ = ∠

So HI is the bisector of angle FHE.

On the other side, 90o

KHI ∠ = hence (KN’FE)=-1

Then (AK,AN’,AF,AE)=-1

But d//BC and MB=MC=1/2BC we get (AK,AM,AB,AC)=-1 or (AK,AN,AF,AE)=-1

Therefore ' N N ≡

Let J be the intersection of AI and BC.

Since BYXC is a cyclic quadrilateral we obtain NXI YXB YCB ICJ ∠ = ∠ = ∠ = ∠



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On the other hand,

902 2 2

o ABC BAC ACB NIX BID IAC ICA JIC

∠ ∠ ∠∠ = ∠ = − = + = ∠ + ∠ = ∠

So NIX JIC △ ∼△ we claim NX JC NI JI

= (1)

Similarly we also have NY JB

NI JI = (2)

From (1) and (2) we obtain . . NX NI JC JI

NI NY JI JB=

Or NX JC AC

NY JB AB= = (q.e.d)

Problem 8 (TTT2-41):

Given a triangle ABC with 60o

A∠ = , the circle (O) inscribed in triangle ABC, touchesthe sides AB,AC,BC at points D,E,F, . The line DE intersects the lines BO, CO,

respectively at N,M. Find the area of triangle MNF in terms of the area of triangle ABC.

Solution:

Case 1: AB=AC. It is easy to prove that SMNP=1/4SABC

Case 2: AB AC ≠

Applying problem 1 we have 90o

BMC ∠ = then two quadrilaterals BDMO and BMOF

are cyclic.



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So NMO ABO CBO FMO∠ = ∠ = ∠ = ∠

Hence NMF B∠ = ∠

Similarly, MNF C ∠ = ∠

Therefore FMN ABC △ ∼△ we claim

2

FMN FMN

ABC ABC

S r

S r

=

(1)

Let H be the projection of O to MF, we have r FMN=OH

Then

2 21

4

FMN

ABC

r OH

r OF

= =

(because

1 130

2 2

o HFO MFN BAC ∠ = ∠ = ∠ = ) (2)

From (1) and (2) we get SFMN=1/4SABC

Problem 9 (TTT2-23):

Let ABC be a non isosceles triangle circumscribed the circle (O).D,E,F are three

tangency points of (O) to the sides BC,CA,AB, respectively. M is the intersection of AO

and DE, N is the intersection of BO and EF, P is the intersection of CO and DF. Showthat SNAB=SMAC=SPBC.

Solution:



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Using problem 1, we have 090 AMB∠ = . Let L be the intersection of BM and AC, K and

H be the projections of M and B to the side AC.

Since BAM CAM ∠ = ∠ and 90o

AMB∠ = we get BM=LM

Then 12

MH BK = or SAMC=1/2SABC

Similarly, SNAB=SPBC=1/2SABC therefore SNAB=SMAC=SPBC (q.e.d)

Comment:

So, starting from a simple geometry problem, we constructed many interesting results.Among them, there are many problems from the contests around the world. I think that

problem 1 still has other applications for the readers to discover.

Date post:	08-Apr-2018
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Simple Problem and Its Applications

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