Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to view more details of problem 1186.
Friday, February 5, 2016
Geometry Problem 1186: Right Triangle, Square, Inscribed Circle, Tangent, Quadrilateral, Area
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http://s17.postimg.org/m0fx94qbz/pro_1186.png
ReplyDeleteDraw circle I similar to problem 1184
Observe that TP tangent to circle O and I
Let 2m and 2n are diameters of circles O and I
Let u= BH and v= BG, a= HT=HM and b= GN=GR
Applying Menelaus theorem in tri. ABC and secant HGP
(BH/HA).(PA/PC).(GC/GB)=1….. (1)
Replace PA/PC=AE/BC= m/n in (1) and simplify it we get
u.v/(v-u)= 2.m.n/(m-n) ……. (2)
Since TP , BC and BA are ext. tangent to circles O and I so we have
Tangent from G to circle O GT=GQ => GH+a= b+NQ….. (3)
Tangent from H to circle I GT=GQ => GH+b= a+MS….. (4)
Subtract (3) to (4) and note that NQ=MS we get a= b.
So HM=HT=GN or v-u= m-n
Replace it in (2) we will have u.v=2.m.n => area of triangle HBG= ½ area of triangle ABC
So line HG bisect triangle ABC into 2 congruent areas
MANOLOUDIS APOSTOLIS
ReplyDelete4 HIGH SCHOOL OF KORYDALLOS -PIRAEUS-GREECE
PROBLEM 1186
Suffice 2.(BHG)=(ABC) or 2.(BG.BH)/2=(AB.BC)/2 or 2.BG.BH=AB.BC.Formed square BCFS end externally tangent circle with center K eat in accordance with the problem 1184 PT
Tangent to the circle with center K. Fetch and second tangent from P intersect the AS to M and DB to L.Is BG=BM(symmetrical with respect to PO).Fetch OQ perpendicular to AB. Is triangles OTH=OHQ Therefore <TOH= <QOH=x. But <BHG=<QOT=2x,<HGB=90-2x.Is <KGP= ½(90+2x)=45=x=<HOK.Therefore OHGK is concycle. Simple OLMK is concycle.But PG.PH=PK.PO=PM.PL therefore O,L,M,K,G,H is concycle.
2.BG.BH=2.BM.BH=2.BK.BO=2.(BC.√(2&2))/2.(AB.√2)/2=AB.BC