Geometry Again!

Following are some geometry questions I couldnot solve. They may be from other olympiads, and if their solutions are available online please point me to them.

  1. For a pentagon ABCDE, AE=ED, AB+CD=BC,ABCDE,\ AE=ED,\ AB+CD=BC, and BAE+CDE=180\angle BAE + \angle CDE=180 ^{\circ} . P.T. AED=2BEC\angle AED=2\angle BEC.

  2. Let P be a point inside a triangle with C=90\angle C=90 ^{\circ} such that AP=ACAP=AC, and let M be the midpt of AB and CH be the altitude. Prove that PM bisects BPH\angle BPH iff A=60\angle A=60 ^{\circ}.

Note by Megh Parikh
5 years, 7 months ago

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This problem wouldn't be so hard once you notice a property that PP carries from AC=APAC=AP.

WLOG say that HH is between A,MA,M. Since AC=AH×ABAC=AH\times AB, therefore AP=AHABAP=AH*AB which is enough to show that APHABPPHPB=APAB=ACAB\triangle APH\sim \triangle ABP\Rightarrow \frac {PH}{PB}=\frac {AP}{AB}=\frac {AC}{AB}. Since PMPM bisects BPH    PHPB=HMBM\angle BPH\iff \frac {PH}{PB}=\frac {HM}{BM} combining the results above we continue the equivalence:     ACAB=HMBM=2HMAB\iff \frac {AC}{AB}=\frac {HM}{BM}=\frac {2HM}{AB}     AC=2HM\iff AC=2HM.

Instead to computing for HMHM, we can geometrically show that AC=2HM    AAC=2HM\iff \angle A. To relate ACAC and HMHM, we introduce the midpoint of ACAC(I actually originally connected GHGH, but then I found a simpler way). Since AGM=CHM=90\angle AGM=\angle CHM=90(CM=AMCM=AM) and AG=AC2=HMAG=\frac {AC}{2}=HM, by HLHL congruence we have GAM=CMA=ACMCAB=60\angle GAM=\angle CMA=\angle ACM\Rightarrow \angle CAB=60. Q.E.D

For the first one I have the same proof as Jorge :)

Xuming Liang - 5 years, 7 months ago

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Thanks.

Megh Parikh - 5 years, 7 months ago

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As a crazy thought for this problem, we could also argue by considering when Apollonius' circle of H,B;MH,B;M and the circle centered at AA with radius ACAC intersect.

Xuming Liang - 5 years, 7 months ago

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For Problem 1. Consider the point FF in the ray BABA such that AF=CDAF=CD (AA between BB and FF), then triangles FAEFAE and CDECDE are congruent (using AE=EDAE=ED and BAE+CDE=180\angle BAE+\angle CDE=180^\circ. Note that the triangles CEBCEB and FEBFEB are congruent because EC=EFEC=EF and BC=AB+CD=AB+AF=BFBC=AB+CD=AB+AF=BF. Finally, BEC=BEF=BEA+AEF=BEA+DEC\angle BEC=\angle BEF=\angle BEA+\angle AEF=\angle BEA+\angle DEC therefore AED=2BEC\angle AED=2\angle BEC.

Jorge Tipe - 5 years, 7 months ago

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Thanks.

Megh Parikh - 5 years, 7 months ago

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