My Math Forum Regular 12 sided polygon

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May 12th, 2018, 11:46 PM   #1
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Regular 12 sided polygon

I have the dilemma with the proof of this problem.

Quote:
 Let $\displaystyle A_{1}A_{2}...A_{12}$ be the regular $\displaystyle 12$ sided polygon. Let $\displaystyle D$ and $\displaystyle E$ be the foots of the altitudes from vertexes $\displaystyle A_{8}$ and $\displaystyle A_{1}$, respectively, of the triangle $\displaystyle A_{1}A_{5}A_{8}$. Let $\displaystyle M$ be the foot of the altitude from point $\displaystyle D$ on $\displaystyle A_{1}A_{8}$. Prove that $\displaystyle A_{1}M=A_{8}E$.
My proof goes like this.

From the regularity of the polygon, we have that $\displaystyle A_{1}A_{8}A_{5}A_{4}$ is an isosceles trapezoid. So, $\displaystyle D$ is point in the intersection of its diagonals and $\displaystyle M$ is the middle point of $\displaystyle A_{1}A_{8}$. It means that $\displaystyle A_{1}M=A_{8}M=DM=EM$ (because $\displaystyle \measuredangle A_{1}DA_{8}= \measuredangle A_{1}EA_{8}=90^{\circ}$).
Also, $\displaystyle \measuredangle MA_{1}D= \measuredangle MA_{8}D=45^{\circ}$ and $\displaystyle \measuredangle A_{4}A_{8}A_{5}= \frac{\frac{360^{\circ}}{12}}{2}=15^{\circ}$.
So, $\displaystyle EM=A_{8}M$ and $\displaystyle \measuredangle MA_{8}E=45^{\circ}+15^{\circ}=60^{\circ}$, which means that $\displaystyle \triangle MA_{8}E$ is equilateral triangle. This means that $\displaystyle A_{8}E=A_{8}M=A_{1}M$.

I am interested in whether I can refer to the regularity of the polygon or I must prove that $\displaystyle A_{1}A_{8}A_{5}A_{4}$ is an isosceles trapezoid?
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Last edited by lua; May 13th, 2018 at 12:12 AM.

 May 13th, 2018, 05:07 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,285 Thanks: 1967 That depends on how formal you want your proof to be. By the way, the reason you gave for $DM = EM$ appears to be insufficient. It becomes sufficient if you also take into account one of the 45° angles you give in your next line. How exactly did you know that $EM=A_8M$? It's an easy consequence of the fact that $E$ lies on the circle with diameter $A_1A_8$, but you seemed to have an even simpler reason in mind.
 May 13th, 2018, 05:47 AM #3 Member   Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2 I want the proof to be as formal as possible. I'll try to explain. From the fact that $\displaystyle A_{1}A_{8}A_{5}A_{4}$ is isoscales trapezoid, I concluded that $\displaystyle M$ is in the middle of $\displaystyle A_{1}A_{8}$, that is, $\displaystyle A_{1}M=A_{8}M$. From the facts that over $\displaystyle A_{1}A_{8}$, $\displaystyle \measuredangle A_{1}DA_{8}=90^{\circ}$ and $\displaystyle \measuredangle A_{1}EA_{8}=90^{\circ}$, I conclude that $\displaystyle A_{8}M=DM=EM$ (that is, as you say, E is on that circle). But I'm worried if it is enough to just say that $\displaystyle A_{1}A_{8}A_{5}A_{4}$ is isoscales trapezoid, and then go on with conclusions. Is it OK or not? Last edited by lua; May 13th, 2018 at 06:07 AM.
 May 13th, 2018, 08:34 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,285 Thanks: 1967 If the proof is allowed to use previously proved theorems, that would depend on whether there was a suitable theorem available about the symmetries of regular polygons.
 May 13th, 2018, 11:57 PM #5 Member   Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2 I guess I'll have to find another way to prove that $\displaystyle M$ is in the middle of $\displaystyle A_{1}A_{8}$.
 May 14th, 2018, 02:34 AM #6 Member   Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2 I'll try to give the proof from the beginning. $\displaystyle A_{5}A_{6}$, $\displaystyle A_{6}A_{7}$ and $\displaystyle A_{7}A_{8}$ are arcs with same length on one circle. So, their central angles are congruent ($\displaystyle \frac{360^{\circ}}{12}=30^{\circ}$), and their inscribed angles are congruent ($\displaystyle \frac{30^{\circ}}{2}=15^{\circ}$). Thus, we may write: $\displaystyle \measuredangle A_{5}A_{1}A_{8}=\measuredangle A_{5}A_{1}A_{6}+\measuredangle A_{6}A_{1}A_{7}+\measuredangle A_{7}A_{1}A_{8}=3 \cdot 15^{\circ}=45^{\circ}$. Now, $\displaystyle \measuredangle A_{1}DM=45^{\circ}$ ($\displaystyle \triangle A_{1}MD$) and $\displaystyle A_{1}M=DM$. Since $\displaystyle A_{8}D \perp A_{1}A_{5}$ and $\displaystyle A_{1}A_{8} \perp DM$, it means that $\displaystyle \measuredangle A_{1}A_{8}D=\measuredangle A_{1}DM=45^{\circ}$. That means that $\displaystyle \triangle DMA_{8}$ is isoscales right triangle with $\displaystyle MA_{8}=DM=A_{1}M$. So, $\displaystyle M$ is in the middle of $\displaystyle A_{1}A_{8}$. Now, points $\displaystyle A_{1}$, $\displaystyle A_{8}$, $\displaystyle E$ and $\displaystyle D$ are on the same circle, so $\displaystyle ME=MA_{8}$, and because $\displaystyle \measuredangle MA_{8}E=60^{\circ}$ (sum of 4 inscribed angles of $\displaystyle 15^{\circ}$), we conclude that $\displaystyle \triangle MA_{8}E$ is equilateral with $\displaystyle A_{8}E=A_{8}M=A_{1}M$. Last edited by lua; May 14th, 2018 at 02:49 AM.
 May 14th, 2018, 11:59 AM #7 Senior Member     Joined: Feb 2010 Posts: 702 Thanks: 137 Since $\displaystyle \angle A_1DA_8 = \angle A_1EA_8 = 90$, that makes $\displaystyle A_1,D,E,A_8$ a cyclic set of points with diameter $\displaystyle A_1A_8$. You have already shown that $\displaystyle \angle DA_1M=45$ so $\displaystyle \triangle A_1DM$ is 45-45-90. Thus $\displaystyle M$ is the center of the circle through the four points and we have that $\displaystyle MA_1=MD=ME=MA_8$. Thanks from lua

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