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A TONCAS that I found

This property is quite well known- If P is a point on the minor arc BC of the circumcircle of an equilateral triangle ABC, then PB+PC=PA.

I discovered the following- If P is a point in the plane of equilateral triangle ABC such that PB+PC=PA, then P is a point on the minor arc BC.

Try to prove this using TONCAS. Hint- A lawyer has a point to make.

Note by Pranav Kirsur
2 years, 6 months ago

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well,I think i haVE GOT ITlet P not be on minor arc BC..take point J in the interior of tr.ABC such that BJ=BP and ang.PBJ=60. so,tr. ABJ is congruent to tr.CBP by SAS congruence criterion..so AJ=PC..JP=BJ=BP(tr.BJP is equilateral)Now consider tr.AJP by triangle inequality we get AP is less than AJ+JP.but AJ+JP=PB+PC...AP=PB+PC...hence contradiction...hence proved!! is this correct?

Neeraja Kirtane - 2 years, 6 months ago

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Yes,this is correct. I like your construction. I used Fermat Point to get a contradiction.

Pranav Kirsur - 2 years, 6 months ago

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What did you do exactly using fermat's point?

Neeraja Kirtane - 2 years, 6 months ago

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@Neeraja Kirtane I have considered a case where P is not on the minor arc BC. I have used the property that the Fermat point minimises the sum of the distances from the vertices of a triangle.

Pranav Kirsur - 2 years, 6 months ago

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@Pranav Kirsur hmm..but how does point P come in this context?

Neeraja Kirtane - 2 years, 6 months ago

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Comment deleted May 21, 2015

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I have already shown it to sir.

Pranav Kirsur - 2 years, 6 months ago

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Try my 1000 followers problem.

Nihar Mahajan - 2 years, 6 months ago

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Ok , then no problem!

Nihar Mahajan - 2 years, 6 months ago

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