Math IOQM

 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...

Q1). A number N in base 10, is 503 in base b and 305 in base b+2. What is the product of the digits of N? S1. N = 248 Answer = 64 Q2).  Let (ABCD) be a rectangle, in which (AB + BC + CD = 20) and (AE = 9), where (E) is the midpoint of the side (BC). Find the area of the rectangle. S2. 2x + y = 20______[1] 81 = x^2 + y^2/4______[2] We just need to find xy. Square the first equation and you will 4x^2 + y^2, replace it using 2 and you will get x.y = 19. Q3). What is the least positive integer by which 2^5. 3^6. 4^3. 5^3.6^7 should be multiplied so that the product is a perfect square? S3. Write it as 2^(5+6+7).3^(6+7).5^3 = 2^18.3^13.5^3 So the answer is 15. Q4). Find the number of integer solutions to ||x| - 2020| < 5 S4. -5 < |x| - 2020 < 5 2015 < |x| < 2025 => x has 18 possible integer solutions when |x| = 2016 to 2024 Q5). The product (55 * 60 * 65) is written as the product of five distinct positive integers. What is the least possible value of the largest of th...

 Q1. ABC is a triangle and D and E are interior points of the sides AB and BC respectively such that: AD/DB = 1/3 CE/EB = 3 If AE and CD intersect at F, find CF/FD. S1. Approach 1: Using mass points : AD/DB = 1/3 => A has higher mass since it's closer to D. Let mA = 3 mB = 1 => mC = 1/3 => mE = 4/3 Now mF = 3 + 4/3 = 13/3 and it matches 4 + 1/3 = 13/3 So we have assigned masses correctly. => CF/FD = 4/(1/3) = 12 Approach 2: Menelaus theorem Typically you would notice that if we have a solution using mass points, we can also solve it using Ceva's theorem or Menelaus' theorem. Here, in triangle BDC, AFE is the traversal intersecting all sides. BE/EC * CF/FD * DA/AB = 1 => 1/3 * CF/FD * 1/4 = 1 => CF/FD = 12 = Answer Q2.  L and M are the mid-points of the diagonals BD and AC respectively of the quadrilateral ABCD. Through D, draw DE equal and parallel to AB. Show that EC || LM EC = 2LM S2. It's straightforward with co-ordinate geometry. A (0,0) B(x1,0) C(...

 Q1. In quadrilateral ABCD, the diagonals AC and BD meet at O. Suppose the four triangles AOB,BOC,COD and DOA are equal in area, prove that ABCD is a parallelogram. S1. [AOB] = [COD] => 1/2.OA.OB.Sin(AOB) = 1/2.OC.OD.Sin(COD) Angles AOB and COD are same. => OA.OB = OC.OD Similarly, OA.OD = OC.OB Multiply both => OA = OC and OB = OD => O bisects both diagonals. If the diagonals bisect each other, the quadrilateral is a ||gram. Why? Let the quadrilateral be A (0,0), B(x,0), C(a,b) D(p,q) O = Midpoint of AC = (a/2,b/2) = Midpoint of BD = ((p+x)/2,q/2) b/2 = q/2 => b = q => CD || AB and a = p+x Slope of AD = q/p Slope of BC = b/(a-x) = q/p => AD || BC Opposite sides are || => ABCD is a ||gram. H.P. Q2. In a parallelogram  A B C D A BC D , a point  P P on the segment  A B A B is taken such that  A P A B = A B A P ​ =   61 2022 2022 61 ​  and a point  Q Q on the segment  A D A D is taken such that  A Q A D = 61 2065...