Problem Relating to Menelaus' Theorem

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bZNyQ7C2

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Hello ...

This is a problem from Purple Comet 2014. Its intent seems to be to give students practice in using the Theorem of Menelaus.

Problem statement: The diagram below shows equilateral triangle ABC with side length 2. Point D lies on ray BC so that CD=4. Points E and F lie on AB and AC, respectively, so that E, F and D are collinear, and the area of triangle AEF is half of the area of triangle ABC. Compute AE.

p 5 Problem 3.png

I produced the diagram myself using Geogebra. I tried to attach my Geogebra file but I couldn't. This site doesn't like the extension. Anyway the only point you can move is E.

I applied the Menelaus' Theorem to triangle ABC with transversal DE. I found this:

\(\displaystyle \begin{array}{c} \frac{{\overline {AE} }}{{\overline {EB} }} \times \frac{{\overline {BD} }}{{\overline {BC} }} \times \frac{{\overline {CF} }}{{\overline {FA} }} = 1\\ \frac{{\overline {AE} }}{{\overline {EB} }} \times \frac{3}{2} \times \frac{{\overline {CF} }}{{\overline {FA} }} = 1\\ \frac{{\overline {AE} }}{{\overline {EB} }} \times \frac{{\overline {CF} }}{{\overline {FA} }} = \frac{2}{3} \end{array}\)

I also applied Menelaus' Theorem to triangle EBC with transversal AC. I learned this:

\(\displaystyle \begin{array}{c} \frac{{\overline {AE} }}{{\overline {AB} }} \times \frac{{\overline {BC} }}{{\overline {CD} }} \times \frac{{\overline {DF} }}{{\overline {FE} }} = 1\\ \frac{{\overline {AE} }}{{\overline {AB} }} \times \frac{1}{2} \times \frac{{\overline {DF} }}{{\overline {FE} }} = 1\\ \frac{{\overline {AE} }}{{\overline {AB} }} \times \frac{{\overline {DF} }}{{\overline {FE} }} = 2\\ \overline {AE} \times \frac{{\overline {DF} }}{{\overline {FE} }} = 4 \end{array}\)

Finally, we're given that the area of triangle ABC is twice that of triangle AEF. That means \(\displaystyle \overline {AE} \times \overline {AF} = 2\).

And of course we have AE plus EB =2 and also AF plus FC = 2. That makes 5 equations with 6 variables.

That's all I know. I have not been able to compute AE from that.

By the way, I do know the length of AE. It's 4/3, and the length of AF is therefore 3/2. I found that by guess work that I checked with measurements made by Geogebra. I just can't derive that number from the problem itself.

I must be missing something. Can anybody give me a hint?
 
I just realized I made a typo on this post. I wrote "applied Menelaus' Theorem to triangle EBC with transversal AC ." I should've written "applied Menelaus' Theorem to triangle EBD with transversal AC ". I hope that didn't confuse anyone.

Still no answers?
 
I assume you have reason to know that Menelaus is intended to be used, rather than just guessing from the appearance of the figure.

My first thought was to work directly with areas. I drew in AD and worked with ratios of areas, then brought in your results and got a quadratic equation whose solutions for AE were 4/3 and -1. My way is probably not the most efficient, but this may give you some ideas.
 
Sorry I'm a little slow replying. I will look at that as soon as I can. Your suggestion sounds promising. It's better than any idea I have at any rate.

I have hesitated to draw in new lines because that would introduce new lengths to measure therefore new variables. For instance, I thought of finding the midpoint of BC and constructing an altitude to A. That would give me three new segments, but I'd know the length of each one. Hmm ... constructing a triangle ABD, its area would be 3 times the area of ABC.

The segment EFD has to be the key to the whole thing. There are many points E and F that will give a triangle half the area of ABC, but only one point E and one point F that will be collinear with D.

I found the problem in a pdf from AOPS by somebody named Michael Tang. It's called "(Ceva's and) Menelaus's Theorem." I'm sure you can find it easily if you are curious. It proves the theorem and offers some practice exercises, including this one. Yes, the theorem of Menelaus is the whole point.
 
@Dr Peterson ... I got the same quadratic equation you did, but maybe not in the same way. I drew in \(\displaystyle \overline {AD} = \sqrt {28} \) and figured a few areas, like \(\displaystyle \left[ {ABC} \right] = \sqrt 3 \) and \(\displaystyle \left[ {ABD} \right] = 3\sqrt 3 \), but without already knowing \(\displaystyle \overline {AE} \) I couldn't prove that \(\displaystyle \left[ {CFD} \right] = \left[ {AEF} \right]\). I'm not sure quite what you did. If you could let me know, I'd appreciate it.

Meanwhile, here's what I did.

\(\displaystyle \begin{array}{c} \frac{{\overline {AE} }}{{\overline {EB} }} \times \frac{{\overline {CF} }}{{\overline {FA} }} = \frac{2}{3}\\ \frac{{\overline {AE} }}{{\left( {\overline {AB} - \overline {AE} } \right)}} \times \frac{{\left( {\overline {AC} - \overline {FA} } \right)}}{{\overline {FA} }} = \frac{2}{3}\\ \frac{{\overline {AE} }}{{\left( {2 - \overline {AE} } \right)}} \times \frac{{\left( {2 - \overline {FA} } \right)}}{{\overline {FA} }} = \frac{2}{3} \end{array}\)

At this point I substituted \(\displaystyle x = \overline {AE} \) and \(\displaystyle y = \overline {AF} \) without forgetting that \(\displaystyle \overline {AE} \times \overline {AF} = xy = 2\). I wrote

\(\displaystyle \begin{array}{c} \frac{x}{{\left( {2 - x} \right)}}\frac{{\left( {2 - y} \right)}}{y} = \frac{2}{3}\\ 3x\left( {2 - y} \right) = 2y\left( {2 - x} \right)\\ 6x - 3xy = 4y - 2xy\\ 6x - 6 = 4y - 4\\ 6x - 2 = 4y\\ 6x - 4y - 2 = 0 \end{array}\)

then used \(\displaystyle xy = 2\) written as
\(\displaystyle y = \frac{2}{x}\)
to eliminate y. The result was

\(\displaystyle \begin{array}{c} 6x - 4y - 2 = 0\\ 6x - 4\left( {\frac{2}{x}} \right) - 2 = 0\\ 6{x^2} - 8 - 2x = 0\\ 3{x^2} - x - 4 = 0\\ \left( {3x - 4} \right)\left( {x + 1} \right) = 0 \end{array}\)

So I got the same result you did but I didn't use any areas.

At this point I doubt there is any easier approach than the one you and I used. Thanks for your help.

.
 
M work was all about ratios of areas, not actually calculating areas. (In fact, I think I entirely neglected to use any given dimensions in that part of the work.) Letting AE = x, I used a sequence of ratios to show that AEF/ABC, which has to be 1/2, is equal to 3*x/2*EF/ED. Then I combined that with your fact from Menelaus that DF/FE = 4/x to make a quadratic equation, which was effectively the same as yours.

Of course, you did use areas to obtain xy = 2, in a way much like what I did.
 
Interesting. There are two Menelaus ratios. I used one and left the other unused. But you used the one I didn't.

I guess this conversation is finished. I wonder if there's a way I can mark the question as answered.
 
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