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May 12th, 2018, 10:46 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?
Attached Images Drz2018_VII_z2.jpg (34.3 KB, 6 views)

Last edited by lua; May 12th, 2018 at 11:12 PM. May 13th, 2018, 04:07 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,746 Thanks: 2133 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, 04:47 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 05:07 AM. May 13th, 2018, 07:34 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,746 Thanks: 2133 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, 10:57 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, 01:34 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 01:49 AM. May 14th, 2018, 10:59 AM #7 Senior Member   Joined: Feb 2010 Posts: 706 Thanks: 141 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 Tags polygon, regular, sided Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Markimus Math 1 May 29th, 2015 05:20 AM Val Bryux Geometry 3 December 23rd, 2014 09:05 AM aditya ranjan Geometry 3 June 16th, 2014 05:09 PM driftaway Algebra 1 June 16th, 2011 07:07 AM quddusaliquddus Algebra 1 May 31st, 2009 03:04 PM

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