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How can I prove this statement? I tried with a pentagon, but I have not achieved anything

Restriction: I can not use the circumference to prove it, and by this I mean to inscribe the polygon in the circumference. The idea is prove it with properties of congruence of triangles or properties of parallelogram or properties of quadrilateral, trapezium or trapezoid.

If all the possible diagonals are drawn in a regular polygon from a vertex, the angles formed in that vertex are equal each.

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  • $\begingroup$ You should clarify what is "allowed". E.g. Can I use vector / complex number geometry (which I'd consider an overkill)? Which properties of regular polygons can I use, especially those about symmetry? $\endgroup$
    – Calvin Lin
    Commented May 4, 2019 at 0:00
  • $\begingroup$ Is there a specific path you would like? E.g. Trigonometry, Coordinate geometry, etc. $\endgroup$
    – Calvin Lin
    Commented May 4, 2019 at 0:03
  • $\begingroup$ The contents that have taught me are: Congruence of triangles, Properties of triangles (median, height, etc), properties of parallelograms, trapezoids and trapezoids. External and internal angles of quadrilaterals and regular polygons. With this I must solve it. $\endgroup$
    – ESCM
    Commented May 4, 2019 at 0:06
  • $\begingroup$ You have to somehow use the fact that all the interior angles from the center to adjacent vertices are equal. That is the definition of regular. $\endgroup$ Commented May 4, 2019 at 0:09
  • $\begingroup$ Mattiu, is Marty's claim allowed? IE Can I assume that a regular polygon has a center, which angle to any given side is equal? $\endgroup$
    – Calvin Lin
    Commented May 4, 2019 at 0:10

4 Answers 4

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Pick a vertex $P_i$ and an adjacent side $P_iP_{i+1}$, and consider the angle bisector of $\angle P_i$ and the perpendicular bisector of $P_iP_{i+1}$.

For each bisector, join opposite vertices that are reflections to each other.

Parallel lines joining some polygon vertices

Do you agree that by symmetry there is a family of parallel lines for each bisector? Then the marked angles are all equal for being alternate angles of parallel lines.

As these parallel lines partition the regular polygon, they form triangles that are congruent to triangles formed by diagonals from a vertex.

e.g. $\triangle P_3P_0P_8 \cong P_0P_3P_4$ by counting the number of vertices between the long edge $P_3P_0$.

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    $\begingroup$ This is very slick. The only proof I could come up with was the one gotten by setting the polygon into a circle, as OP forbade. $\endgroup$
    – Lubin
    Commented May 4, 2019 at 3:15
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How about this? But, as you can see, it's skirting about the "construct circle". In particular, the claim can be a step in proving "Angle at the center is twice angle at circumference".

Assumption: A regular polygon has a center $O$, where for any consecutive vertices $B, C$, $\angle BOC$ is a constant $ \frac{360^\circ}{n }$.

Claim: In triangle ABC, if $OA=OB=OC$, then $\angle BAC = \frac{1}{2} \angle BOC$.
This is easily proven using isosceles triangles and angle chasing.

Hence, the result follows.


I hope this approach doesn't "use circumference"

Hint: A regular polygon can be inscribed in a circle.

Hint: Angle at the center is twice angle at circumference. Clearly, the angles at center are all the same.


I'm guessing this approach is "using circumference"?

Hint: A regular polygon can be inscribed in a circle.

Hint: If $A, B, C$ are 3 points on a circle, then $\angle ABC $ is dependent only on the length $AC$.

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  • $\begingroup$ You are solving it with the circumference and its angles, I can not do that. $\endgroup$
    – ESCM
    Commented May 3, 2019 at 23:53
  • $\begingroup$ I think you're using the circumscribed polygon concept and $ OA = OB = OC $ are the radius of the circle and therefore are equal, but I can not use that, the idea is to solve it with the concepts I mentioned, without any circumference concept. $\endgroup$
    – ESCM
    Commented May 4, 2019 at 0:23
  • $\begingroup$ While I understand your frustration of "I cannot use X", you need to be much clearer in specifying what is allowed or not allowed. Am I allowed to assume that a regular polygon has a center which satisfies OA=OB=OC? If no, then 1) What is your definition of a regular polygon, and 2) Why can't I use that definition to show that such an O exists? $\endgroup$
    – Calvin Lin
    Commented May 4, 2019 at 0:26
  • $\begingroup$ Thanks for your good help, I did not know that by DEFINITION that was true? $\endgroup$
    – ESCM
    Commented May 4, 2019 at 0:29
  • $\begingroup$ So the definition of regular polygon, is given by the circumference? $\endgroup$
    – ESCM
    Commented May 4, 2019 at 0:30
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Let the regular polygon be $P_0P_1P_2\cdots P_{n-1}$ where $P_i$'s are vertices in order. The polygon has $n$ sides where $n\ge 4$ to be interesting.

Let each interior angle be $\theta$:

$$\theta = 180^\circ-\frac{180^\circ\cdot2}{n}$$

Pick $P_0$ as the vertex to draw diagonals from.

Consider the diagonal $P_0P_2$. The triangle cut off $\triangle P_0P_1P_2$ is isosceles with $P_0P_1 = P_1P_2$. The base angles are

$$\angle P_1P_0P_2 = \angle P_1P_2P_0 = \frac{180^\circ - \theta}2 = \frac{180^\circ}{n}$$


Now here's an inductive assumption: The $i$th triangle formed by $\triangle P_0P_iP_{i+1}$ has the following angles:

$$\begin{align*} \angle P_iP_0P_{i+1} &= \frac{180^\circ}n\\ \angle P_iP_{i+1}P_0 &= \frac{180^\circ \cdot i}n \end{align*}$$


For the next (i.e. the $(i+1)$th) triangle $\triangle P_0P_{i+1}P_{i+2}$, reflect $P_{i}$ by diagonal $P_0P_{i+1}$ to obtain $P_{i}'$. $P_{i}'$ may be outside or inside or on the edge of the polygon, but the cases are similar.

Examples of $i=1$ and $i=5$ in a regular nonagon:

Example of vertex reflection with a nonagon n = 9

In both cases, the proof strategy is to:

  1. Consider the polygon interior angle $\angle P_iP_{i+1}P_{i+2}$ and three angles sharing the same vertex: $\angle P_iP_{i+1}P_0$, $\angle P_0P_{i+1}P_i'$ and $\angle P_i'P_{i+1}P_{i+2}$;
  2. Use the inductive assumption to solve $\angle P_i'P_{i+1}P_{i+2}$;
  3. Consider the isosceles triangle $\triangle P_i'P_{i+1}P_{i+2}$, and solve its base angles $\angle P_{i+1}P_i'P_{i+2}$ and $\angle P_i'P_{i+2}P_{i+1}$;
  4. Show that $P_0$, $P_i'$ and $P_{i+2}$ are collinear using the first base angle;
  5. Prove the induction statement for $i+1$.

If $P_{i}'$ is inside the polygon, e.g. $i=1$ above, consider the triangles at $P_{i+1}$: $\triangle P_0P_iP_{i+1}$, $\triangle P_0P_i'P_{i+1}$ and $\triangle P_i'P_{i+1}P_{i+2}$.

$$\begin{align*} \angle P_iP_{i+1}P_{i+2} &= \angle P_iP_{i+1}P_0 + \angle P_0P_{i+1}P_i' + \angle P_i'P_{i+1}P_{i+2}\\ 180^\circ - \frac{180^\circ\cdot2}{n} &= \frac{180^\circ \cdot i}n + \frac{180^\circ \cdot i}n + \angle P_i'P_{i+1}P_{i+2}\\ \angle P_i'P_{i+1}P_{i+2} &= 180^\circ - \frac{180^\circ (2i+2)}{n} \end{align*}$$

$P_{i+1}P_{i+2} = P_{i+1}P_i'$ for being the sides of the regular polygon, so $\triangle P_i'P_{i+1}P_{i+2}$ is isosceles. Consider its base angles:

$$\begin{align*} \angle P_{i+1}P_i'P_{i+2} &= \frac{180^\circ - \angle P_i'P_{i+1}P_{i+2}}{2}\\ &= \frac{180^\circ(i+1)}n\\ &= \angle P_i'P_{i+1}P_0 + \angle P_i'P_0P_{i+1} \end{align*}$$

And so $P_0P_i'P_{i+2}$ is a straight line, so

$$\begin{align*} \angle P_{i+1}P_0P_{i+2} &= \angle P_{i+1}P_0P_i'\\ &= \frac{180^\circ}n\\ \angle P_{i+1}P_{i+2}P_0 &= \angle P_{i+1}P_{i+2}P_i'\\ &= \frac{180^\circ(i+1)}n \end{align*}$$


If $P_i'$ is outside the polygon, e.g. $i=5$ above, consider at $P_{i+1}$,

$$\begin{align*} \angle P_iP_{i+1}P_{i+2} &= \angle P_i P_{i+1} P_0 + \angle P_0 P_{i+1} P_i' - \angle P_i'P_{i+1}P_{i+2}\\ 180^\circ - \frac{180^\circ\cdot2}{n} &= \frac{180^\circ \cdot i}n + \frac{180^\circ \cdot i}n - \angle P_i'P_{i+1}P_{i+2}\\ \angle P_i'P_{i+1}P_{i+2} &= -\left[180^\circ - \frac{180^\circ (2i+2)}{n}\right] \end{align*}$$

$\triangle P_i'P_{i+1}P_{i+2}$ is again isosceles with $P_{i+1}P_{i+2} = P_{i+1}P_i'$. Consider its base angles:

$$\begin{align*} \angle P_{i+1}P_i'P_{i+2} &= \frac{180^\circ - \angle P_i'P_{i+1}P_{i+2}}2\\ &= 180^\circ - \frac{180^\circ(i+1)}n\\ &= \angle P_0P_i'P_{i+1} \end{align*}$$

So in this case $P_0P_{i+2}P_i'$ is a straight line, and so

$$\begin{align*} \angle P_{i+1}P_0P_{i+2} &= P_{i+1}P_0P_i'\\ &= \frac{180^\circ}n\\ \angle P_{i+1}P_{i+2}P_0 &= 180^\circ - P_{i+1}P_{i+2}P_i'\\ &= \frac{180^\circ(i+1)}n\\ \end{align*}$$


By induction, all $\angle P_iP_0P_{i+1}$ are equal to $\frac{180^\circ}n$ for $i = 1, 2, \ldots , n-2$.

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Since the polygon is regular, it means that all the angles of interest (made up by the lines drawn from a single point) lie on a same-length segment of the circumscribed circle (equal sides of the regular polygon) and hence must be the same. However, I'm not sure if this answer doesn't satisfy your 'circumference' requirement (please clarify if it doesn't).

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  • $\begingroup$ I cant inscribe the polygon in a circumference to prove it $\endgroup$
    – ESCM
    Commented May 3, 2019 at 23:55

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