Synthetic Geometry Session Followup Questions and Problem Summary

In our first session, we spent some time on this problem: Let ABCABC be a triangle in which A=60\angle A = 60^{\circ} Let BEBE and CFCF be the angle bisectors with EE on ACAC and FF on ABAB. Let MM be the reflection of AA in the line EFEF. Prove that MM lies on BCBC

We discovered a few key properties about this configuration that eventually led to a successful solution:

  1. AEIFAEIF is cyclic because BIC=90+A2=120    FIE+A=180\angle BIC=90+\frac {\angle A}{2}=120^{\circ}\implies \angle FIE+\angle A=180^{\circ}. The cyclic result only occurs when A=60\angle A=60^{\circ}, and it gives us information about the angles of AEF\triangle AEF and consequently those of FME\triangle FME

  2. II is the circumcenter of MEFMEF because IE=IFIE=IF and FIE=2FME\angle FIE=2\angle FME. This result relates MM with II and gives us more angle relations.

  3. BMIFBMIF is cyclic, which we proved by using the above properties to angle chase FBI=FMI\angle FBI=\angle FMI. Symmetrically CMIECMIE is cyclic, so IMB+IMC=AFI+AEI=180\angle IMB+\angle IMC=\angle AFI+\angle AEI=180^{\circ}, which implies B,M,CB,M,C are collinear and we are done.

The discovery of these properties makes the problem transparent, meaning we know what makes our main result true. What this also means is that we can probably find simpler solutions. Indeed, there are many ways to solve this problem:

Using angle bisector theorem and Miquel point (Avoids tedious angle chasing):

Since AEIFAEIF is a cyclic quadrilateral, we know the miqual point of its complete correspondence lies on BCBC. This means the circumcircles of BFI,CEI,ABE,ACFBFI,CEI, ABE,ACF all have a common point on BCBC, we denote the point MM'. Because ACMFACM'F is cyclic, FMBF=ACBC=AFBF\frac {FM'}{BF}=\frac {AC}{BC}=\frac {AF}{BF}; therefore AF=FMAF=FM' and similarly AE=EMAE=EM'. This is enough to establish M=MM=M' and MM lies on BCBC

Utilizing EFEF as the axis of symmetry:

Let EFBC=PEF\cap BC=P. It suffices to prove PEPE bisects APB\angle APB. We can achieve this with our first property.

Now I will propose a few follow up questions which you guys should now easily answer:

We keep the notations of our main problem:

  1. DBCDBC is equilateral such that D,AD,A are on opposite sides of BCBC. Prove that DE=DFDE=DF

  2. Through II construct the perpendicular to EFEF which intersects EF,BCEF,BC at X,YX,Y. Prove that 2IX=IY2IX=IY

  3. OO is the circumcenter of AEFAEF, Prove that OMBCOM\perp BC.

Note by Xuming Liang
5 years, 8 months ago

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For follow up problem 1 :

To prove DE=DFDE=DF , it suffices to establish congruence between ΔDIF\Delta DIF and ΔDIE\Delta DIE (not dead yet) . We have DI=DIDI=DI and we have already established that IF=IEIF=IE by having IFE=IEF=30\angle IFE= \angle IEF=30. So we need to prove a pair of angles congruent. Since we want a good connection between the angles and the discovered stuff , we prefer proving DIF=DIE\angle DIF=\angle DIE. By cyclicity we have mBIF=mCIE=60m\angle BIF=m\angle CIE=60. So we need to prove DIB=DIC=BIC2=1202=60\angle DIB=\angle DIC=\dfrac{\angle BIC}{2}=\dfrac{120}{2}=60. Now how to achieve this? We have not yet utilized the equilateral triangle yet!

So, consider BICD\Box BICD Can we prove it cyclic? Yes! We have to prove that EBD+ECD=180\angle EBD+\angle ECD=180 and its easy.

IBD+ICD=IBC+CBD+ICD+BCD=B2+60+C2+60=120+60=180BICD is cyclic! \angle IBD+\angle ICD =\angle IBC+\angle CBD+\angle ICD+\angle BCD=\dfrac{B}{2}+60+\dfrac{C}{2} + 60=120+60=180 \\ \Rightarrow \Box BICD \ is \ cyclic!

So by cyclicity we have DIB=DIC=60\angle DIB=\angle DIC=60 which we had to prove.

Nihar Mahajan - 5 years, 8 months ago

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Very nice reasoning using congruence! For the cyclic part, I believe you meant to say B,I,C,DB,I,C,D are concyclic, which is true because BIC+BDC=180\angle BIC+\angle BDC=180. Since BD=DCBD=DC, IDID bisects BIC\angle BIC, which is what you wanted to prove.

Xuming Liang - 5 years, 8 months ago

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Sorry , yes it is that B,I,C,DB,I,C,D are concyclic. I have edited it and few angles too.

Nihar Mahajan - 5 years, 8 months ago

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A fast way to prove that BICDBICD is cyclic is to notice that BIC=120\angle BIC = 120^{\circ} and BDC=60\angle BDC = 60^{\circ} .

Alan Yan - 5 years, 7 months ago

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For follow up problem 2:

First we have ΔIXF\Delta IXF as 30-60-90 triangle. So FI=2IXFI=2IX. So we have to prove that FI=IYFI=IY. So consider ΔBIF\Delta BIF and ΔBIY\Delta BIY , where we have FBI=YBI\angle FBI=YBI , BI=BIBI=BI , BIF=BIY=60\angle BIF=\angle BIY=60 (by angle chase). So they are congruent by ASA test and hence FI=IYFI=IY is proved.

Nihar Mahajan - 5 years, 8 months ago

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I pretty much proved it the same as you. Nice solution.

Sharky Kesa - 5 years, 8 months ago

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For follow up problem 3:

Yesterday we did find that EMC=60\angle EMC=60 So to prove our result , we must prove that OME=30\angle OME=30. Using the fact that OO is circumcentre , we have EOF=EIF=120\angle EOF=\angle EIF=120 and by some angle chase we have OEI=OFI=60\angle OEI=\angle OFI=60 where OFE=IFE=OEF=IEF=30\angle OFE=\angle IFE=\angle OEF=\angle IEF=30. We get a beautioful result that FOEI\Box FOEI is a rhombus which has one of its diagonals that is OIOI equal to its side. Using these facts and that II is circumcenter of ΔEMF\Delta EMF we have EI=OI=IF=IMEI=OI=IF=IM and thus EOFM\Box EOFM is cyclic giving OFE=OME=30\angle OFE= \angle OME=30 which was to be proved :)

Nihar Mahajan - 5 years, 8 months ago

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When will be the next discussion?

Shivam Jadhav - 5 years, 8 months ago

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Problem 3: Since MM is the Miquel point, there exists a spiral similarity centered at MM that maps FIAEFI \rightarrow AE. Denote M1,M2M_1, M_2 as the midpoint of AFAF and EIEI, respectively. Thus, the spiral similarity maps FIM1M2FI \rightarrow M_1M_2. Using the circumcircles of BFI\triangle BFI and BM1M2BM_1M_2, this implies that M1M2BMM_1M_2BM is cyclic. Since M1OM2BM_1OM_2B is cyclic, this implies that M1OM2MBM_1OM_2MB is cyclic with diameter OBOB. Thus, OMB=90\angle OMB = 90^{\circ} and we are done.

Alan Yan - 5 years, 7 months ago

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Great solution. This fact is actually true as long as BICDBICD is cyclic.

Xuming Liang - 5 years, 7 months ago

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