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This is a tough program I have hard time to find the answer. Can anyone help me? Thank you very much in advance.

Problem - In regular pentagon $ABCDE$, point $M$ is the midpoint of side $AE$, and segments $AC$ and $BM$ intersect at point $Z$. If $ZA = 3$, what is the value of $AB$? Express your answer in simplest radical form.

This is a drawing for the problem:

enter image description here

Théophile
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  • An illustration would be extremely useful here (and would also prove some effort on your side, which this question currently lacks evidence of). – barak manos Mar 11 '16 at 14:00
  • How do I add a drawing after posting the problem? I'm new here. Thank you. –  Mar 11 '16 at 14:05
  • Click edit, and then click the image icon at the top of your post. – barak manos Mar 11 '16 at 14:09
  • I wonder whether there is a simple to solve the problem. This is a middle school problem. –  Mar 11 '16 at 15:11
  • A nice solution can be found from this address: https://www.mathcounts.org/sites/default/files/u5328/2015%20State%20Competition%20Solutions%20-%20FINAL.pdf –  Mar 11 '16 at 16:43

2 Answers2

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Choose the coordinate system so that $A = (0,0)$, $B = (\rho,0)$ and the other three vertices lie in the upper half plane. Identify the plane with complex numbers, the five sides of the pentagon corresponds to $$( AB, BC, CD, DE, EA ) \quad\leftrightarrow\quad (\rho,\rho\omega,\rho\omega^2,\rho\omega^3,\rho\omega^4)$$ where $\omega = e^{2\pi i/5}$ is the primitive $5^{th}$ root of unity.

Under this identification,

$$B = AB = \rho,\; C = AB+BC = \rho(1 + \omega) \quad\text{ and }\quad M = -\frac12 EA = -\frac{\rho}{2} \omega^4 = -\frac{\rho}{2} \bar{\omega}$$

Since $Z$ lies on the intersection of $AC$ and $BM$, there exists real numbers $\lambda, \mu$ such that $$AZ = \lambda AC = (1-\mu) AB + \mu AM$$ This impies $$\frac{Z}{\rho} = \lambda( 1 + \omega) = (1-\mu) - \mu\frac{\bar{\omega}}{2} = \left[1 - \mu \left( 1 + \frac{\omega + \bar{\omega}}{2} \right)\right] + \frac{\mu}{2}\omega $$ The square bracket in RHS is clearly a real number. By comparing the real and imaginary part of the last equality, we get:

$$\lambda = 1 - \mu\left(1 + \frac{\omega + \bar{\omega}}{2} \right) = \frac{\mu}{2} \implies \lambda = \frac{1}{3 + \omega + \bar{\omega}} $$

This implies

$$|AB| = \rho = \left|\frac{AZ}{\lambda(1 + \omega)}\right| = 3 \left|\frac{3+\omega+\bar{\omega}}{1+\omega}\right| = 3\left(\frac{3 + 2\cos\frac{2\pi}{5}}{2\cos\frac{\pi}{5}}\right) = 3\left(\frac{2+\phi}{\phi}\right) = 3\sqrt{5} $$

achille hui
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    Thank you very much, Achille. I wonder whether there is a simple to solve the problem. This is a middle school problem. –  Mar 11 '16 at 15:12
  • @teamember all the steps above can be rephrased in turns of vectors. The proof will be more ugly than this but it is doable by middle schoolers. There should be more ingenious way to get the answer but I haven't figure it out yet. – achille hui Mar 11 '16 at 15:19
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Let $F$ be the intersection of $AC$ and $BE$. By angle chasing, we obtain the following picture:

enter image description here

Now $\triangle BAF$ and $\triangle BEA$ are similar, so $$BF:BA = BA:BE,$$ or $$\frac yx = \frac{x}{x+y}=\frac{1}{1+y/x}.\tag{1}$$ Solving that for $y$ in terms of $x$ we have $$\frac yx = \frac{\sqrt{5}-1}2.$$

Now apply Menelaus' Theorem to $\triangle AEF$ with $M$, $Z$, $B$ colinear we get $$\frac{EM}{MA}\cdot \frac{ZA}{ZF}\cdot \frac{BF}{BE} = 1.\tag{2}$$ Note that $EM = MA$, $$\frac{BF}{BE} = \frac{y}{x+y}=1-\frac{x}{x+y}=\frac{3-\sqrt{5}}2,$$ and $$\frac{ZA}{ZF} = \frac{3}{y-3}.$$ Thus, (2) becomes $$\frac{y-3}{3} = \frac{3-\sqrt{5}}2.$$ So $$ y = \frac{9-3\sqrt{5}}{2}+3 = \frac{15-3\sqrt{5}}2.$$ And we get $$ x = \frac xy \cdot y = \frac{\sqrt{5}+1}2 \cdot \frac{15-3\sqrt{5}}2$$

Note:

  1. I'm sure there are more elegant solutions.
  2. One can avoid Menelaus' Theorem by connecting $M$ with the midpoint $N$ of $AF$.
Quang Hoang
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