## Friday, August 11, 2017

### Geometry Problem 1342: Circle, Secant, Chord, Midpoint, Concyclic Points, Cyclic Quadrilateral

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

1. https://goo.gl/photos/GUjUkpRUbFxGVfvd8

1. Consider triangles FHA and AHE
Since AC//GD so ∠ (HAF)= ∠ (AGD)= ∠ (HED)= u
Triangle FHA similar to AHE ( case AA)
So HA/HE= HF/HA = > HA ^2= HE.HF= HB.HC= HN^2
Since H is the midpoint of NA => A,N,B,C form a Harmonic conjugated points i.e (A,N,B,C)= (C,B,N,A) =- 1
2. Since M is the midpoint of BC so MB^2=MC^2= MN. MA ..( properties of Harmonic conjugated points)
Draw new circle with BC as a diameter and from A draw a tangent to this circle at T ( see sketch)
Let N’ is the projection of T over MA
Relation in right triangle MTA give MT^2= MN’.MA= MB^2= MN. MA => N’ coincide to N
Relation in right triangle MTA give AT^2= AM.AN= AB.AC = AD.AE ( power from A to circles M and O)
So D,E, M, N are cocyclic

2. Problem 1342
Is <HFA=<GFE=<GDE=<HAE so triangle AHF is similar with triangle EHA so AH^2=HF*HE=HB*HC=HN^2.From Newton (A,N,B,C)=-1.
So AB/BN=AC/CN or AC*BN=AB*CN or AC*(AN-AB)=AB*(AC-AN) or
AC*AN-AC*AB=AB*AC-AB*AN or 2AB*AC=AN*(AC+AB) or
Therefore MNDE is cyclic.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE

3. Problem 1342

Need a bit of algebraic simplification in addition to geometry.

Let MN = u, NB = v and BH = w so that
MC = u+v and AH = v+w.

Now < HAF = < FGD = < FED and so
AH is a tangent to circle AFE.

Hence (v+w)2 = HF.HE = w(2u+2v+w) which simplifies to v2 = 2uw ….(1)

= (v+2w)(2u+3v+2w)
= 2(v + w)(2u + 3v + 2w) – v(2u + 3v + 2w)
= 2(v + w)(u + 2v + 2w) + 2(v + w)( u + v) – v(2u + 3v + 2w)
= 2(v + w)(u + 2v + 2w) + (2uv + 2uw + 2v2 + 2vw – 2uv – 3v2 – 2vw)
= 2(v + w)(u + 2v + 2w) + (2uw – v2)
= 2(v + w)(u + 2v + 2w) from (1)
= AN.AM

M,N,D,E are concyclic

Sumith Peiris
Moratuwa
Sri Lanka

4. Extend EN to cut the circle O at P. Let GP meet AC at M'.
<FED=<FGD=, by AC||GD,=<HAF.
So triangles HEA and HAF are similar so HA^2=HF*HE=HN^2.
Since HN^2=HF*HE, triangles HNF and HEN are similar so <HNF=<NEH.
But <NEH=<PGF so <HNF=<NEH=<PGF so GFNM' is cyclic so AN*AM'=AF*AG=, by power of point in circle O,=AD*AE.
So AN*AM'=AD*AE so EDNM' is cyclic so <NED=<NM'D=, by AC||GD,=<GDM'.
But <NED=<PGD so <NED=<GDM'=<PGD so GM'D is isosceles so M' is in fact the midpoint of BC so M'=M so cyclic=EDNM'=EDNM.