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Synthetic Geometry Session Followup Questions and Problem Summary

In our first session, we spent some time on this problem: Let \(ABC\) be a triangle in which \(\angle A = 60^{\circ}\) Let \(BE\) and \(CF\) be the angle bisectors with \(E\) on \(AC\) and \(F\) on \(AB\). Let \(M\) be the reflection of \(A\) in the line \(EF\). Prove that \(M\) lies on \(BC\)

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

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

  2. \(I\) is the circumcenter of \(MEF\) because \(IE=IF\) and \(\angle FIE=2\angle FME\). This result relates \(M\) with \(I\) and gives us more angle relations.

  3. \(BMIF\) is cyclic, which we proved by using the above properties to angle chase \(\angle FBI=\angle FMI\). Symmetrically \(CMIE\) is cyclic, so \(\angle IMB+\angle IMC=\angle AFI+\angle AEI=180^{\circ}\), which implies \(B,M,C\) 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 \(AEIF\) is a cyclic quadrilateral, we know the miqual point of its complete correspondence lies on \(BC\). This means the circumcircles of \(BFI,CEI, ABE,ACF\) all have a common point on \(BC\), we denote the point \(M'\). Because \(ACM'F\) is cyclic, \(\frac {FM'}{BF}=\frac {AC}{BC}=\frac {AF}{BF}\); therefore \(AF=FM'\) and similarly \(AE=EM'\). This is enough to establish \(M=M'\) and \(M\) lies on \(BC\)

Utilizing \(EF\) as the axis of symmetry:

Let \(EF\cap BC=P\). It suffices to prove \(PE\) bisects \(\angle APB\). 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. \(DBC\) is equilateral such that \(D,A\) are on opposite sides of \(BC\). Prove that \(DE=DF\)

  2. Through \(I\) construct the perpendicular to \(EF\) which intersects \(EF,BC\) at \(X,Y\). Prove that \(2IX=IY\)

  3. \(O\) is the circumcenter of \(AEF\), Prove that \(OM\perp BC\).

Note by Xuming Liang
1 year, 10 months ago

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For follow up problem 3:

Yesterday we did find that \(\angle EMC=60\) So to prove our result , we must prove that \(\angle OME=30\). Using the fact that \(O\) is circumcentre , we have \(\angle EOF=\angle EIF=120\) and by some angle chase we have \(\angle OEI=\angle OFI=60\) where \(\angle OFE=\angle IFE=\angle OEF=\angle IEF=30\). We get a beautioful result that \(\Box FOEI\) is a rhombus which has one of its diagonals that is \(OI\) equal to its side. Using these facts and that \(I\) is circumcenter of \(\Delta EMF\) we have \(EI=OI=IF=IM\) and thus \(\Box EOFM\) is cyclic giving \(\angle OFE= \angle OME=30\) which was to be proved :) Nihar Mahajan · 1 year, 10 months ago

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For follow up problem 2:

First we have \(\Delta IXF\) as 30-60-90 triangle. So \(FI=2IX\). So we have to prove that \(FI=IY\). So consider \(\Delta BIF\) and \(\Delta BIY\) , where we have \(\angle FBI=YBI\) , \(BI=BI\) , \(\angle BIF=\angle BIY=60\) (by angle chase). So they are congruent by ASA test and hence \(FI=IY\) is proved. Nihar Mahajan · 1 year, 10 months ago

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@Nihar Mahajan I pretty much proved it the same as you. Nice solution. Sharky Kesa · 1 year, 10 months ago

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For follow up problem 1 :

To prove \(DE=DF\) , it suffices to establish congruence between \(\Delta DIF\) and \(\Delta DIE\) (not dead yet) . We have \(DI=DI\) and we have already established that \(IF=IE\) by having \(\angle IFE= \angle IEF=30\). 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 \(\angle DIF=\angle DIE\). By cyclicity we have \(m\angle BIF=m\angle CIE=60\). So we need to prove \(\angle DIB=\angle DIC=\dfrac{\angle BIC}{2}=\dfrac{120}{2}=60\). Now how to achieve this? We have not yet utilized the equilateral triangle yet!

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

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

So by cyclicity we have \(\angle DIB=\angle DIC=60\) which we had to prove. Nihar Mahajan · 1 year, 10 months ago

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@Nihar Mahajan A fast way to prove that \(BICD\) is cyclic is to notice that \(\angle BIC = 120^{\circ}\) and \(\angle BDC = 60^{\circ} \). Alan Yan · 1 year, 9 months ago

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@Nihar Mahajan Very nice reasoning using congruence! For the cyclic part, I believe you meant to say \(B,I,C,D\) are concyclic, which is true because \(\angle BIC+\angle BDC=180\). Since \(BD=DC\), \(ID\) bisects \(\angle BIC\), which is what you wanted to prove. Xuming Liang · 1 year, 10 months ago

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@Xuming Liang Sorry , yes it is that \(B,I,C,D\) are concyclic. I have edited it and few angles too. Nihar Mahajan · 1 year, 10 months ago

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Problem 3: Since \(M\) is the Miquel point, there exists a spiral similarity centered at \(M\) that maps \(FI \rightarrow AE\). Denote \(M_1, M_2\) as the midpoint of \(AF\) and \(EI\), respectively. Thus, the spiral similarity maps \(FI \rightarrow M_1M_2\). Using the circumcircles of \(\triangle BFI\) and \(BM_1M_2\), this implies that \(M_1M_2BM\) is cyclic. Since \(M_1OM_2B\) is cyclic, this implies that \(M_1OM_2MB\) is cyclic with diameter \(OB\). Thus, \(\angle OMB = 90^{\circ} \) and we are done. Alan Yan · 1 year, 9 months ago

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@Alan Yan Great solution. This fact is actually true as long as \(BICD\) is cyclic. Xuming Liang · 1 year, 9 months ago

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When will be the next discussion? Shivam Jadhav · 1 year, 10 months ago

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