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What have you tried?
Dr.Peterson
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There are several ways; try expressing each marked angle in terms of a corresponding central angle, and think about how those add up.
What have you learned about such figures?
What have you learned about such figures?
shahar
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I understand that it is equal 180 degree because every angle is half central angle and the sum is 360 degree divide by 2.
Can you explain the way of Aion?
Dr.Peterson
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I don't know what it is.
I can add that several methods for variations on this problem can be found on my site:
Angles in a Star – The Math Doctors
www.themathdoctors.org
Most of this is about the harder problem that doesn't assume a regular polygon (but which also applies to that case); but one picture deals with your special case using my method.
My method was the following. I connected the vertices to form an inscribed pentagon. Then I used the inscribed angle theorem to determine all of the inscribed angles. Then I summed these, which turned out to be equal to [imath]3(\alpha + \beta+\delta+\gamma +\lambda)[/imath]. Now recall that the sum of the interior angles in an [imath]n[/imath]-sided polygon is [imath](n-2)\pi[/imath]. Hence for [imath]n=5[/imath] we have [imath]\alpha + \beta+\delta+\gamma +\lambda =\pi[/imath].
Remains to support the claim that the (outer) angle of the pentagon equals three times the angle of the pentagram.
Not sure what you mean. I showed that the sum of the interior angles for the inscribed pentagon was equal to [imath]3(\alpha + \beta+\delta+\gamma +\lambda)[/imath]. But we know that the sum of the interior angles is equal to [imath]3\pi[/imath] for a pentagon. Hence [imath]3(\alpha + \beta+\delta+\gamma +\lambda)=3\pi[/imath]. Divide by 3 and we obtain [imath](\alpha + \beta+\delta+\gamma +\lambda)=\pi[/imath].
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I meant that I understand the [imath] \alpha [/imath] at the top but not why the other two occurrences are equal to [imath] \alpha, [/imath] too. If so, then the rest is clear. For symmetry reasons, we have [imath] \alpha=\beta=\gamma=\delta=\lambda [/imath]. That means that you say that the angle of the pentagram is a third of the angle of the pentagon. Why is it?
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The way I obtained the other two occurrences of [imath]\alpha[/imath] for example was by using a corollary to the inscribed angle theorem. Namely, that inscribed angles that stand on the same arc are equal. I'm not sure how to prove I that [imath] \alpha=\beta=\gamma=\delta=\lambda [/imath].
Yes. The result of my "proof" is that the sum of the interior angles of the pentagram is a third of the sum of the interior angles of the pentagon. Why this is true on a deeper level I'm not sure. I just brute-forced a solution without thinking that deeply.
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Let's see. We have
for symmetry reasons, since any rotation about [imath] 72° [/imath] is congruent, and thus the equations [imath] \alpha +2 \beta =0.6 \,\pi [/imath] from the angle in a pentagon, and [imath] \alpha +4\beta=\pi [/imath] from the triangle [imath] \triangle ABC. [/imath] The difference is [imath] 2\beta = 0.4\,\pi [/imath] and so [imath] \beta=0.2\,\pi. [/imath] Hence [imath] \alpha=0.6\,\pi - 2\cdot 0.2\,\pi= 0.2\,\pi=\beta .[/imath]
for symmetry reasons, since any rotation about [imath] 72° [/imath] is congruent, and thus the equations [imath] \alpha +2 \beta =0.6 \,\pi [/imath] from the angle in a pentagon, and [imath] \alpha +4\beta=\pi [/imath] from the triangle [imath] \triangle ABC. [/imath] The difference is [imath] 2\beta = 0.4\,\pi [/imath] and so [imath] \beta=0.2\,\pi. [/imath] Hence [imath] \alpha=0.6\,\pi - 2\cdot 0.2\,\pi= 0.2\,\pi=\beta .[/imath]
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Now it makes sense! One angle of the pentagram is a third of the angle of the pentagon, therefore also the sum
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