Good proof problems

1) if S is the circumcentre of a \(\triangle ABC\) Then prove that \(\angle BSD = \angle BAC\) given that D is the mid point of BC.

2) Let P be any point inside a regular polygon. If did_i is the distance of P from the ithi^{th} side, prove that d1+d2+d3+....+dnd_1+d_2+d_3+....+d_n is constant.

3) I is the incentre of Triangle ABC.X and Y are the feet of the perpendiculars from A to BI and CI. Prove XY || BC

I liked these problems, so i thought that I should share them with the community! :D

Edit: Xuming Liang is forbidden to comment.

Note by Mehul Arora
4 years, 11 months ago

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1 vote

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Question 3: Denote the point where BIBI hits ACAC as EE. Denote the angle measures as A,B,CA, B, C according to the vertices of the triangle. All angles in this solution are in degrees.

By Exterior Angle Theorem, we have that BEA=C+B2\angle BEA = C + \frac{B}{2} . Since AYE=90\angle AYE = 90, we have that YAE=90CB2\angle YAE = 90 - C - \frac{B}{2} .

Connecting AA to II, we know that AIAI is the angle bisector of A\angle A, which implies that IAE=90B2C2\angle IAE = 90 - \frac{B}{2} - \frac{C}{2} . This implies that IAY=C2\angle IAY = \frac{C}{2} .

Now notice since AXI=AYI=90\angle AXI = \angle AYI = 90, we have that quadrilateral AXIYAXIY is cyclic. This means that if we connect XX and YY, we have that IAY=IXY=C2\angle IAY = \angle IXY = \frac{C}{2} . However we also know that XCB=C2\angle XCB = \frac{C}{2} . Therefore by alternate interior angles, we know that XYBCXY || BC.

Alan Yan - 4 years, 11 months ago

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@Alan Yan How u so genius?

Mehul Arora - 4 years, 10 months ago

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Question 2. Partition the polygon into triangles. Let mm be the side length and let AA be the total area. The area of the triangles will be 12md1,12md2,12md3,... \frac{1}{2}md_1 , \frac{1}{2}md_2 , \frac{1}{2}md_3 , ... . Adding these areas yield the total area.

This implies that 12m(d1+d2+...+dn)=A    di=2Am\frac{1}{2}m(d_1+d_2+...+d_n) = A \implies \sum{d_i} = \frac{2A}{m} which means the sum of the distances is constant.

Alan Yan - 4 years, 11 months ago

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Hints: 1. Property of circumcenter 2. Area

Xuming Liang - 4 years, 11 months ago

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You are not allowed to comment on such posts :) :P

Nihar Mahajan - 4 years, 11 months ago

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Exactly :P

Mehul Arora - 4 years, 11 months ago

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@Xuming Liang Geom god! _/_

The first one is quite simple. The second one is interesting :)

But you man! These probs are a left hand's play for you, provided you are right handed. :3

Mehul Arora - 4 years, 11 months ago

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in triangle BDC and CDS,



and angle d is 90

so both triangles are congruent and then angle BSD = angle CSD

also 2.angleBSD = 2.angleBAC

hence proved

Dev Sharma - 4 years, 11 months ago

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Question 1. BSD=12BSC\angle BSD = \frac{1}{2}\angle BSC (Perpendicular Bisector.)

BAC=12BSC\angle BAC = \frac{1}{2}\angle BSC (Central angle and inscribed angle substending the same arc.)

Therefore BSD=BAC \angle BSD = \angle BAC.

Alan Yan - 4 years, 11 months ago

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