Math IOQM

 

  Key point is to convince yourself that each vertex of the square is also the center of its nearby circular arc. Once you do that you can see that area of the shaded region  = a^2 - 4*1/4*pi*(a/2)^2 = a^2(1 - pi/4). Now how to convince yourself? Notice that the arcs SR and RQ are touching each other at the point R which is the midpoint of CD. So they share a common tangent passing through R. Also SR and RQ arcs are mirror images of each other along the vertical line passing through R. If they are mirror images of each other touching each other at R and the mirror is placed vertically at R then that mirror itself is the tangent. If mirror is the tangent then the radius at that point will be perpendicular to it, and that will lie on the side CD. Only point on CD which is equidistant from R and Q is C, hence that's the center. 

  If P is a point dividing segment AB in ratio m:n and it lies between A and B like this: A____(m)____P____(n)____B Then position of P will be (m*B + n*A)/(m+n). For e.g. A is 0 and B is 9. P divides AB into 2:1. Then P = (1*0 + 2*9)/(2+1) = 18/3 = 6. A___(6)___P___(3)___B But if P lies outside AB to the right of B, like this: A_______(9)___B_____(9)____P P still divides AB into 2 : 1 since AP/BP = 18/9 = 2/1. But we can't apply our earlier formula correctly. So we say that P divides AB into 2: -1 or -2: 1. Now if we apply the formula: P = (2*9 + -1*0)/(2-1) = 18 OR P = (-2*9 + 1*0)/(-2 + 1) = 18 If P lies to the left of A then it can't divide AB into 2: -1 since AP will always be less than BP. Now it can divide AB into -1:2 or 1:-2 So P = (-1*9+2*0)/(2-1) = -9 P____(-9)_____A______(9)____B

 Rectangle properties: Problems: Prove that medians/altitudes/angle bisectors of a triangle are concurrent. For altitudes: Solution: If you apply the formula for CosA and CosC this problem is effectively to prove that c.CosA = b.CosC. but c.CosA = AH So pr. that AH = b.CosC but CosC = CH/a So pr. that. AH/CH = b/a By Angle Bisector theorem b/a = AD/BD. So pr. that AH/CH = AD/BD. By Ceva's theorem AD/BD * BM/CM * CH/AH = 1 But BM = CM So AD/BD = AH/CH Hence proved.

  Question 1 Solution: ACQD is ||gram coz AD || CQ and AD = CQ. So [DPQ] = [APC] from congruence. [APC] = [BPC] coz same base(PC) and height from opposite side. So [DPQ] = [BPC]. Question 2 Solution: [LZY] = [XYZ] + [TXZ] + [LTX] [MZYX] = [XYZ] + [TXZ] + [MTZ] So we need to prove that [LTX] = [MTZ]. LMZX is a trapezium so triangles between parallel sides are similar and other 2 have equal area which can be proven using side ratios. Which gives [LTX] = [MTZ]. I am not sure what was the use of LX = XY = YN here? Question 3 Solution: [ADF] = [BDF] = 3 coz same base (DF) and height. [FCE] = [ADF]  = 3due to AAS congruence. (AD = CE, F is vertical angle, angle DAF = angle CEF). [FCB] = [FCE] = 3 coz same base(BC = EC) and height (from F to BE). Diagonal BD splits ||gram ABCD in 2 equal areas, prove by congruence. So [BDF] + [BFC] = 1/2[ABCD] = 3 + 3 So [ABCD] = 12. Question 4 Solution : Can be easily proven that [RDF] = [RAS] = [BSP] = [PFC] = k/8 where k  = [ABCD]. Hint: ASR ...

  You can say that BD = 0 in right angle triangle. Solution: First try to solve it using a square of side 1. You will quickly figure out that area of all parts of square except AXY adds up to 5/8 so AXY's area will be 3/8. So p + q = 11. Now try to do it using a ||gram. Let's say sides of ABCD are a,b. In triangle BCD, area is k/2 where k = [ABCD], i.e. area of ABCD. Triangle CXY is similar to BCD with side ratio 1/2 so area ratio = 1/4. [CXY] = k/8. One important thing in ||gram is that adjacent angles are supplementary and opposite angles are equal. So sin(x) for each angle is same. [CXY] = 1/2(a/2)(b/2)sin(x) = k/4 => ab.sin(x) = 2k [ADY] = ab/2.1/2.sin(x) = k/4 = [ABX] So [AXY] = k - k/4 - k/4 - k/8 = 3k/8 p + q = 11. Solution: angle ACE is right angle since sides of ACE make a Pythagorean triplet. [ABF] = 1/2*3*15*sin(A), sin(A) = 16/20 [ABF] = 18 Similarly [EFD] = 5*12*1/2*sin(E) = 18 [BCD] = 1/2*9*4 = 18 [ACE] - [BFD] = [ABF] + [EFD] + [BCD] = 54 1 - 1/sqrt(2) Solutio...