Math IOQM

Q0. In a ||gram ABCD, does the diagonal AC bisect the angle DAB? Q1. M,N,D are midpoints of AB,AC,BC. X,Y are midpoints of NC,ND. Find [MNXY]/[ABC]. S1. Let [ABC] = f => [AMN] = f/4 => [BCNM] = 3f/4 [CND] = f/4 => [NYX] = f/16 [MNXY] = [NYX] + [MNY] => we need to find [MNY] now. [BDNM] = [BCNM] - [NDC] = f/2 [BDM] = [NDM] = f/4 [MNY] = [DMY] = f/8 => [MNXY] = [NYX] + [MNY] = f/16 + f/8 = 3f/16 Q2. ABCD is a parallelogram. X divides AB in the ratio 3:2. Y divides CD in the ratio 4:1. XY intersects AC at Z. Find (AZ : AC). S2.

 1. Show that length of direct common tangent of 2 non intersecting circles(touching is fine) is sqrt(d^2 - (r1-r2)^2) and that of the transverse tangent is sqrt(d^2 - (r1+r2)^2) where 'd' is the distance between the centers of those 2 circles. 2. Now show that if the 2 circles touch each other than DCT length = 2.sqrt(r1r2). 3.  If two circles with radii (a) and (b) touch each other externally. Let (c) be the radius of a circle that touches these two circles as well as a common tangent of these two circles. Prove 1/sqrt(c) = 1/sqrt(a) + 1/sqrt(b)

 Q1. In △ABC, AD, BE and CF are concurrent lines. P, Q, R are points on EF, FD, DE such that DP, EQ and FR are concurrent. Prove that AP, BQ and CR are also concurrent. S1. Prerequisite: Standard and trigonometric form of Ceva's theorem. Standard form gives us: AF/FB * BD/DC * CE/EA = 1 and EP/PF * FQ/QD * DR/RE = 1 Using the trigonometric form, we need to prove that: Sin(CAP)/Sin(PAB) * Sin(ABQ)/Sin(QBC) * Sin(BCR)/Sin(RCA) = 1 Look at triangles EAP and PAF and compute their area ratio: [EAP]/[PAF] = AE * AP * Sin(CAP)/AF*AP*Sin(PAB) = EP/PF Similarly do for others. Multiply both sides of all 3. You will get your desired relation. H.P. Q2. If (X) and (Y) are variable points on the sides (CA, AB) of (\triangle ABC) such that CX/XA + AB/AY = 1 prove that (XY) passes through a fixed point. S2. CX/XA > 0 since everything is positive. => AB/AY < 1 => AY > AB => Y lies on AB extended towards B. => AB = AY - BY => AB/AY = 1 - BY/AY From here: we have 2 cases for X...

 Q1. The internal angle bisector of ∠A of △ABC meets BC at P and (b = 2c) in the usual notation. Prove that ((9AP^2 + 2a^2)) is an integral multiple of (c^2). S1. Let PC = n, BP = m, AP = d We need to prove: 9d^2 + 2a^2 = k.c^2 where 'k' is an integer. c/m = b/n by Angle Bisector theorem. => 2m = n => 3m = a => m = a/3, n = 2a/3 Using Stewart's theorem : man + dad = bmb + cnc = m.4c^2 + nc^2 = a(mn + d^2) = c^2 (4m + n) => a(n^2/2 + d^2) = c^2(3n) => a(4a^2/18 + d^2) = c^2.2a/3 => 4c^2/3 = (2a^2/9 + d^2) => 12c^2 = (2a^2 + 9d^2) H.P. Q2. In (\triangle ABC), in the usual notation, the area is bc/2 sq. units. (AD) is the median to (BC). Prove that Angle ABC = (Angle ADC)/2 S2. Area = 1/2.bc.sin(A) => A = 90 deg. So D is the circumcenter. Chord AC subtends angle ADC at center and ABC at circumference. H.P. Q3. Let C1 be any point on (AB) of triangle ABC. Draw CC1 meeting AB at C1. The lines through A and B parallel to CC1 meet BC produced and AC produce...

Q1) Let Γ be a semicircle with diameter AB. The point (C) lies on the diameter (AB) and points (E) and (D) lie on the arc (BA), with (E) between (B) and (D). Let the tangents to (\Gamma) at (D) and (E) meet at (F). Suppose that (\angle ACD = \angle ECB). Prove that (\angle EFD = \angle ACD + \angle ECB). S1: Let's quickly get an intuition about it. If the point C is same as the center then CDFE is a cyclic quadrilateral since the angles at D and E are both 90. Angle at C is 180 - 2.theta so the angle at F is 2.theta which is what we need to show. H.P. So it works for the case when the point C is same as the center. Now, let's try the general case. Now ODFE is the cyclic quadrilateral. => Angle DFE = 180 - DOE So proving that DFE = ACD + ECB = 2.theta is same as proving that: DOE = 180 - 2.theta = DCE But if we need to prove that DOE = DCE then we need to prove that E,D,C,O are concyclic since the chord DE is subtending the same angles at O,C. Now look at the original problem...

Q1. The diagonals of a parallelogram (ABCD) intersect at (O). A line through (O) intersects (AB) at (X) and (DC) at (Y). Another line passing through (O) intersects (AD) at (P) and (BC) at (Q). Prove that (XQYP) is a parallelogram. S1. Triangles OAX and OCY are congruent. Why? OA = OC Angle CYO = AXO Angle AOX = YOC => OX = OY => O bisects XY Similarly O bisects PQ. So in quadrilateral XQYP, the diagonals bisect each other. Hence it's a ||gram. H.P. So it gives us a property about ||gram that a line passing through diagonal midpoint and ending on opposite sides is also bisected at the diagonal midpoint. Q2. Prove that the feet of perpendiculars drawn from the vertices of a parallelogram onto its diagonals are the vertices of another parallelogram. S2. Let the feet of perpendiculars from D,B on AC be Q,P. We will try to show that OP = OQ so that we can show that O bisects PQ. Similarly, O will also bisect the other line joining the other 2 feet of perpendiculars. And in that c...