Math IOQM

Proof: Compute ma and mb using appollonius and equate. 2[ma^2 + a^2/4] = b^2 + c^2 __[1] 2[mb^2 + b^2/4] = a^2 + c^2 __[2] ma^2 - mb^2 = (b^2 - a^2)/2 - (a^2 - b^2)/4 = 3/4[b^2 - a^2] H.P.

Yes. Why? In first triangle: a/SinA = 2R Next triangle: p/SinP = 2R Angles are same => sides are same H.P.

Q1. In a convex quadrilateral ABCD , angle ABD = 30 , angle BCA = 75 , angle ACD = 25 , and CD = CB . Extend CB  to meet the circumcircle of triangle DAC  at E . Prove that CE = BD . S1. Q2. Let ABC be a triangle with circumcircle Gamma, and let G be the centroid of triangle ABC. Extend AG, BG, and CG to meet the circle Gamma again at A1, B1, and C1, respectively. Suppose that triangle ABC ~ A1B1C1, specifically with: angle BAC = angle A1B1C1,  angle ABC = angle A1C1B1,  and angle ACB= B1A1C1. Prove that ABC and A1B1C1 are equilateral triangles. S2. Step 1. Do angle chasing. Same chord same angle. Step 2. Use given angle equalities. A1 = C => 5 + 6 = 4 + 6 => 4 = 5 B1 = A => 1 + 4 = 1 + 2 => 4 = 2 C1 = B => 3 + 5 = 3 + 2 => 2 = 5 Now draw the modified figure: And all the medians intersect at Centroid G. We know that medians divide the triangle into 3 equal areas. [GAB] = [GBC] = [GCA] Each of these triangles as angle 2. [GAB] = GB.AB.sin(2)/2 [GBC] ...

Q1. Let ABCDE be a paper pentagon with AB = AE, angle A = angle B = angle E = 90, BC = 3, CD = 5, and DE = 2. Construct a perpendicular from  A to CD using only a ruler and drawing not more than six lines. All lines have to be drawn inside the pentagon. S1. Make A = (0,0) assign co-ordinates and you can get radius r = 6. Also the circle centered at A with radius AB = r will be touched at B, the tangent from C to B is of length 3. Similarly the tangent DE = 2. So the other tangents from D,C to the same circle will have lengths 2,3 respectively. Using the right triangles ABC, AED, we can show that AC = 3.sqrt(5) and AD = 2.sqrt(10) Now let K be the point on CD s.t. CK = 3 and KD = 2 (such a point will exist since CD = 5). In triangle ACD, apply stewart's theorem to get AK = 6. So, point K will also lie on the circle. And CD is the tangent to it. Triangle ABC and AKC are congruent. So angle ACK = ACB, hence CA is angle bisector of angle BCD. Similarly AED,AKD are congruent and DA is a...

In a trapezoid ABCD, A + D = 180 since AB || CD. Since it's cyclic, B + D = 180. => A = B and C = D  But a trapezoid with 2 equal base angles has the non parallel sides equal in length. Why? Drop perpendiculars from A,B to side CD at E,F AE = BF since they are distances between the 2 parallel sides. Since C = D angles => ADE and BCF are congruent => AD = BC So ABCD being cyclic => it's isosceles. H.P.

 Q1. Let ABCD be a convex quadrilateral with ∠BAD=2∠BCD and AB=AD. Let P be a point such that ABCP is a parallelogram. Prove that CP=DP. S1. Draw the ||gram first. That will make it easier to draw the rest of the diagram. Let O be center of circumcircle of BCD BOD is 2.theta. OB = OD = r Given AB = AD = x. => AO is perpendicular bisector of BD. Now we can show BOD and BAD are congruent triangles. BD common. Both are isosceles so AO is angle bisector of 2.theta for both. Angle ABD = ADB = OBD = ODB = 90 - theta By ASA congruent. So corresponding sides are equal. => r = x. BODA is Rhombus. Now CP is also r cause ABCP is a ||gram. Since BODA is Rhombus AB || OD. Also ABCP is ||gram so AB || CP => AB || OD || CP Also AB = CP = r = OD OD = CP and OD || CP => ODPC is a ||gram. => DP = OC = r H.P. that CP = DP. Q2. The ratio of the median AM of a triangle ABC to the side BC equals sqrt(3):2 The points on the sides of ABC dividing these sides into 3 equal parts are marked. P...