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Consider the following figure:

enter image description here

$\Delta DAC, \Delta CEB, \Delta AFB$ are isosceles. $\angle ADC = \angle CEB = \angle AFB = 120^{\circ}$.

Prove that $\Delta DEF$ is equilateral.

Now, there is a proof for this using coordinate geometry. And its an ugly one. I was wondering if there was a 'better' proof for this using affine geometry.

We are required to prove that:

$$||\vec{DE}|| = ||\vec{EF}|| = ||\vec{FD}||$$

$$\implies (\vec{E} - \vec{D})\cdot (\vec{E} - \vec{D}) = (\vec{F} - \vec{E})\cdot (\vec{F} - \vec{E}) = (\vec{D} - \vec{F})\cdot (\vec{D} - \vec{F})$$

$$\implies 2\cdot\vec{E} = \vec{F} + \vec{D}$$

$$\implies 2\cdot\vec{D} = \vec{F} + \vec{E}$$

$$\implies 2\cdot\vec{F} = \vec{D} + \vec{E}$$

How should I proceed?

EDIT: Thanks for all the wonderful solutions below, but I require a solution through affine geometry only.

Gerard
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  • What is your intuition in using affine geometry only? Angles and equilateral triangles are not natural in affine geometry. It would seem to me that you would lose more than you would gain by using affine trickery. – Kieran Cooney Dec 05 '13 at 04:39
  • Then what is natural in affine geometry? – Gerard Dec 09 '13 at 09:58

3 Answers3

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This is how I look at the problem:

enter image description here

Since $\overline{AO_1}=\overline{CO_1}$, $\overline{AO_2}=\overline{BO_2},$ and $\overline{BO_3}=\overline{CO_3}$, we can draw circles with centers $O_1,\ O_2,$ and $O_3$ intersecting points $A,C$; $A,B$; and $B,C$, respectively. Now, the first thing we notice is that all three of these circles intersect at a point $D$. How do we prove that?

Let $D_1$ be the intersection of $O_1$ and $O_2$ that is not $A$, $D_2$ be that of $O_1$ and $O_3$ that is not $C$, and $D_3$ be that of $O_2$ and $O_3$ that is not $B$. Note that $\angle AO_1 D_1=2\angle ACD_1$. Meanwhile $\angle D_1O_2B=2\angle D_1AB=2\angle D_1 AC$. Note that $\angle D_1AC+\angle D_1CA+ \angle CD_1A=180^\circ$. We then have $\frac{1}{2}\angle D_1 O_2 B+\frac{1}{2}\angle D_1 O_1 A+ (360-2\angle AO_1C)=180^\circ$, which ultimately simplifies to $\angle D_1O_2B+\angle D_1O_1A=120^\circ$. Similarly, we obtain $\angle D_2O_1C+\angle D_2O_3B=\angle D_3O_2B+\angle D_3O_3B=120^\circ$.

Therefore $\angle D_1O_1A+\angle CO_1D_2+\angle BO_2D_1+\angle D_2O_3B=240^\circ$ or $\angle BO_2D_1+\angle D_2O_3B=120^\circ-\angle D_2O_1D_1$ which ultimately implies $\angle D_2O_1D_1+\angle D_1O_2D_3+\angle D_3O_3D_2=0$. Since $D_1, D_3,$ and $A$ lie on $O_2$ and $D_1, D_2$, and $A$ lie on $O_1$, $2\angle D_2AD_1+2\angle D_1AD_3+\angle D_3O_3 D_2=2\angle D_2AD_3+\angle D_3O_3D_2=0$. However, this implies $A$ lies on the circle with center $O_3$ unless $D_2=D_3$, therefore $D:=D_1=D_2=D_3$.

Now for the fun part. Note that $\overline{O_1O_2}$ is an angle bisector of $\angle DO_1A$ and $\angle AO_2D$ since the quadrilateral $AO_2DO_1$ forms a kite. Similarly $\overline{O_1O_3}$ bisects $\angle CO_1D$ and $\angle DO_3C$ and $\overline{O_2O_3}$ bisects $\angle BO_2D$ and $\angle DO_3B$. So $$\angle O_3O_1O_2=\angle O_3O_1D+\angle DO_1O_2=\frac{1}{2}(\angle DO_1A+\angle CO_1D)=\frac{1}{2}(120^\circ)=60^\circ$$ $$ \angle O_2O_3O_1=\angle DO_3O_1-\angle DO_3O_2=\frac{1}{2}(\angle DO_3C-\angle DO_3B)=\frac{1}{2}(120^\circ)=60^\circ $$ $$ \angle O_1O_2O_3=\angle O_1O_2D-\angle O_3O_2D=\frac{1}{2}(\angle AO_2D-\angle BO_2D)=\frac{1}{2}(120^\circ)=60^\circ $$

$\Rightarrow$ $\triangle O_1O_2O_3$ is equiangular

Note: Sorry for all the angle bash!

Tim Ratigan
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0

A hint: Show that the red and the blue triangles are congruent.

enter image description here

Christian Blatter
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Hint: Draw equilateral triangles at $A$ and $B$: enter image description here

newzad
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