# A Problem with a Heptagon

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.

Note by Daniel Liu
4 years, 9 months ago

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From 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.

- 4 years, 9 months ago

Because any isosceles trapezoid is cyclic.

- 4 years, 9 months ago

Alternately, since the original heptagon is cyclic.

- 4 years, 9 months ago

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

- 4 years, 8 months ago

well I know that too,still its not the exact one.

- 4 years, 8 months ago

its from B.MATH HONS. ADM TEST 2009 BY ISI

- 4 years, 9 months ago

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.

- 4 years, 9 months ago

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

- 4 years, 9 months ago

please do.i have no fb accnt

- 4 years, 9 months ago

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.

- 4 years, 8 months ago

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 :)

- 4 years, 9 months ago