Math IOQM

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...

PHP1: n holes, n+1 pigeons, at least 1 hole with > 1 pigeons. PHP2: n holes, nk+1 pigeons, at least 1 hole with > k pigeons. Q1. Pr. th. If there are 5 points inside a unit square, then the distance between at least any two points is less than 1/sqrt(2). A1. Divide in 4 smaller squares => at least 2 will be inside one square. Max distance inside that square is 1/sqrt(2). Q2. There are 14 sock pairs in a basket. How many you need to pick to be sure that you have 1 pair at least. A2. 15. PHP3: If mean of n +ive integers is k then at least 1 of them is >= k and at least 1 of them is <= k. Proof: If a1 = a2 = ... an = k then H.P. If a1 <= a2 <= a3 ... <= an then if a_i > k => a_j < k H.P. Q3. Suppose numbers from 1 through 20 are placed in some order around a circle. Prove that i) The sum of some three consecutive numbers must be at least 32. ii) The sum of some four consecutive numbers must be at least 42. Let the triplets be (a1,a2,a3) (a2,a3,a4) ... (a20...