Prove that in a regular heptagon $V_1V_2\cdots V_7$, that $\dfrac{1}{V_1V_2}=\dfrac{1}{V_1V_3}+\dfrac{1}{V_1V_4}$.

Challenge: Prove it only using standard Geometry.

I know this problem is from a contest somewhere, but I can't find the source; it would be nice if someone can post a link to the source.

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## Comments

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TopNewestFrom Ptolemy on $V_1 V_2 V_3 V_5$, we have $V_1 V_2 \cdot V_3 V_5+V_2 V_3 \cdot V_1 V_5=V_1 V_3 \cdot V_2 V_5$.

Because $V_2 V_5=V_1 V_4=V_1 V_5$, $V_3 V_5=V_1 V_3$, and $V_2 V_3=V_1 V_2$ the equation can be rewritten as $V_1 V_2 \cdot V_1 V_3+V_1 V_2 \cdot V_1 V_4=V_1 V_3 \cdot V_1 V_4$

Dividing by $V_1 V_2 \cdot V_1 V_3 \cdot V_1 V_4$ on both sides, we have $\frac{1}{V_1V_2}=\frac{1}{V_1V_3}+\frac{1}{V_1V_4}$, as desired.

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Because any isosceles trapezoid is cyclic.

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Alternately, since the original heptagon is cyclic.

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Its from my facebook page.You may see it. Its link is https://www.facebook.com/StudyingMath . And a solution using trignometry is given there.This question is somewhere in the last ones.Check it.And if u don't have fb account, Then I can post that solution here.

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please do.i have no fb accnt

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Sorry for the late reply.I was engaged with my school exams.Here is a link for the imahe.If you still have any difficulties, then you can tell me. https://imageshack.com/i/141fnrp . Sorry I can't upload an image on brilliant.And let me know if you did not get any steps.And after the image It follows that 7y/4=(pi)/2 so y=2(pi)/7. therefore number of sides =7.

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No, sorry; I'm pretty sure it is from a competition. Please do post the solution here, though.

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it`s from B.MATH HONS. ADM TEST 2009 BY ISI

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The problems come from Indian INMO 1992. See here : Its problem 9.

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well I know that too,still it`s not the exact one.

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An announcement here: sorry for not posting any #CosinesGroup posts recently. I have a bunch of homework recently, so I didn't find the time to write any.

I promise that the next post will come no later than a week :)

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