Geometry Problem

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

Click the figure below to see the problem 728 details.

## Saturday, February 11, 2012

### Problem 728: Triangle, Altitude, Perpendicular, Circumcircle, Midpoint, Intersecting Circles, Concurrency

Labels:
altitude,
circumcircle,
concurrent,
intersecting circles,
midpoint,
perpendicular,
triangle

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Note that centers O and O’ will be on AC.

ReplyDeleteLet NG cut EF at L and HM cut EF at L’

NG Perpen. To AB & HM perpen. To BC

Quadrilaterals ANQL and CDL’M are cyclic

Power from point P to circle ANQL= BN.BA=BL.BD

Power from point P to circle CDL’M=BM.BC=BL’.BD

Power of point P to circles O and O’ =BN.BA=BE.BF=BM.BC

From above we have BL.BD=BL’.BD => BL=BL’ => L coincide to L’

So BF, GN and MH are concurrent

My solution

ReplyDeletehttp://lectiimatematice.blogspot.com/2012/02/problem-728-triangle-altitude.html

A beautiful application of Ceva's Theorem (Trigonometric Version)

ReplyDeleteSketch:

Let EAC = x, ECA = y

AE, BE, CE are concurrent Cevians satisfying

sin ABD.sin ECB.sin EAC

= sin EBC.sin ECA.sin EAB

which implies

cos A.sin(C-y).sin x = cos C.sin y.sin (A-x)...(#)

EF,GN,HM are Cevians thro'the vertices E,G,H of triangle EGH.

sin HED=sin HCE=sin y ;

sin EGN=sin EAN=sin(A-x);

sin CHM=sin(90-C)=cos C

while

sin DEG=sin GAE=sin x;

sin NGA=sin(90-A)=cos A;

sin MHE=sin MCE= sin(C-y)

Clearly in view of (#)

The product of the above first three ratios

= the product of the above next three ratios.

Hence by Ceva, EF,GN,HM are concurrent.

Problem 728

ReplyDeleteThe ΒF is the radical axis of centers of circles O and O ',then BN.BA=BE.BF=BM.BC. So

A,N,M,C are concyclic.Then <MNB=<BCA.But If I bring the altitudes ck and AL (T = orthocenter).Is <AKC=<ALC=90 then A,K,L,C are concyclic. So <LKB=<BCA.Therefore

KL//NM then BK/BN=BL/BM ,but if NG and HM intersect the BF to P, P ' respectively

(KT//PN,LT//MP’). Is BK/BN=BT/BP, BL/BM=BT/BP’. So BT/BP=BT/BP’ then BP=BP’.

Therefore NG,MH,BD are concyrrent.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE