practice problems pending

 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 ABCD, a point P on the segment AB is taken such that APAB= 612022 and a point Q on the segment AD is taken such that AQAD=612065. If PQ intersects AC at T find ACAT to the nearest integer.

S3.



To find AC/AT we can look at triangles formed using AC and AT and relate their areas.

[APQ] = 1/2.AP.AQ.sin(QAP)
[ABD] = 1/2.AB.AD.sin(DAB), angle DAB = QAP

[APQ]/[ABD] = AP/AB.AQ/AD = 61^2/2022.2065_________[1]

[APQ] can also be written as [APT] + [AQT]

[APT]/[ABC] = AP/AB * AT/AC_______[2]
[AQT]/[ADC] = AQ/AD * AT/AC_________[3]

But [ABC] = [ADC] = [ADB]
So [APT] + [AQT] = [ABC] * AT/AC[AP/AB + AQ/AD] = [ABC].61^2/2022.2065
Using 1,2,3.
=>
AT/AC[61/2022 + 61/2065] = 61^2/2022.2065
=>
AT/AC = 61/(2022+2065)
AC/AT = 4087/61 = 67 = Answer


Q3.



S3.

AC is the common diagonal for both.
AC^2 = 9 + 121 = 63 + 81 = 130

For finding the area of shaded ||gram we need to find its base since height = AB = 3.
If FC and AD intersect at P, then we need to find AP.
FAP is a right triangle with one side known.
But angle FAP = FAC - DAC and if we know angle FAP then we can find all the sides of triangle FAP.

Let angle FAC = x
tan(x) = CF/AF = 9/7

Let angle DAC = y
tan(y) = CD/AD = 3/11

If FC and AD intersect at P then AFP is a right angle triangle.
And angle FAP = x - y
tan(x-y) = tan(x) - tan(y)/1 + tan(x).tan(y) = 9/7 - 3/11/ 1 + 27/77
= 78/104 = 3/4
=> cos(FAP) = 4/5 = AF/AP = 7/AP
=> AP = 35/4

Area of ||gram shaded = 35*3/4 = 105/4 = m/n
m + n = 109 = answer

Q4.
In △ABC, the straight lines AD,BE,CF are drawn through a point P to meet BC,CA,AB at D,E,F respectively. Prove that PD/AD+PE/BE+PF/CF=1 and AP/AD+BP/BE+CP/CF=2.

S4.
[PBC]/[ABC] = PD/AD
[PAC]/[ABC] = PF/CF
[PAB]/[ABC] = PE/BE
Add all:
([PAC] + [PAB] + [PBC])/[ABC] = [ABC]/[ABC] = PD/AD + PF/CF + PE/BE H.P.

Now AP/AD = (AD-PD)/AD = 1 - PD/AD
Similarly other ratios.
Add all to get 3 - 1 = 2
H.P.

Q5. In quadrilateral ABCD, if angle bisectors of A and C meet at BD then prove that angle bisectors of B and D meet at AC.
S5.
Let angle bisectors of A and C meet at O which lies on BD.
Then in triangle ABD, AB/AD = BO/OD
In triangle BCD, BC/CD = BO/OD
=> AB/AD = BC/CD______[1]

Let angle bisector of B intersect AC at P.
=> AB/BC = AP/PC

Let angle bisector of D intersect AC at Q.
=> AD/DC = AQ/QC

If P,Q were the same point(which we need to prove), then:
AD/DC = AB/BC
but it's already true from [1]
which means that P,Q are indeed same.
which means that Angle bisectors of B and D intersect AC at the same point.
which means that they meet at AC.
H.P.

Q6. ABCD is a trapezium with AB∥CD and AB=2CD. If the diagonals meet at O, then prove that 
3AO=2AC. If AD and BC meet at F, then prove that AD=DF.
S6.

AOB and COD are similar with ratio 2:1

so AO = 2OC

=> AC = 3OC

=> 3AO = 2AC = 6OC

H.P.


DFC and AFC are similar with ratio 1:2 with DC || AB

=> DC is the midline of triangle || to AB

so D is midpoint of AF => AD = DF. H.P.



Q7.



S7.
In a trapezium, diagonals bisect each other in the ratio of || sides.

In ABCD,
AP:PC = AD:BC
=> AD = 7

In ABCE,
AP:PC = AB:CE
=> CE = 26

Also, BP:PE = 1:2 and BP:PD = 2:1
DE = PE - PD = 2BP - 0.5BP = 1.5BP

Can we find BP now?
In triangle ABC, we know all sides and a cevian BP divides AC in 2 known segment lengths.

So we can use Stewart's theorem.
man + dad = bmb + cnc
5.10.15 + 15.BP^2 = 5.14^2 + 10.13^2
=> BP^2 = 128
=> DE = 1.5.sqrt(128) = answer

Comments

Post a Comment

Popular posts from this blog

IOQM 2024 Paper solutions (Done 1-21, 29)

Image
Q1 (number-theory). The smallest positive integer that does not divide 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 is: Solution : This product has all the integers from 1 to 9. So smallest integer not dividing it would be the next prime after 9. Which is 11. Answer: 11 Q2 (combinatorics): The number of four-digit odd numbers having digits 1, 2, 3, 4, each occurring exactly once, is: Solution: Last digit has to be 1 or 3: 2C1 Choosing remaining 3 digits: 3C3 Arranging those 3 digits: 3! Answer: 2C1*3C3*3! = 12 Q3 (number-theory): The number obtained by taking the last two digits of \(5^{2024}\) in the same order is: Solution: To get last 2 digits from number, divide by 100 and take the remainder. So we have to compute mod 100. 25 mod 100 = 25 25.5 mod 100 = 125 mod 100 = 25 5^3.5 mod 100 = 25.5 mod 100 = 25 5^4.5 mod 100 = 25.5 mod 100 = 25 ..... So: 5^2024 mod 100 = 25. Q4 (geometry): Let ABCD be a quadrilateral with \(\angle ADC = 70^\circ\), \(\angle ACD = 70^\circ\), \(\angle ACB = 10^\circ\...
Read more

Simon's factoring trick(complete the rectangle)

  "Complete the Rectangle" , also called Simon’s Favorite Factoring Trick , is a clever algebraic method for factoring expressions of the form: x y + a x + b y + c xy + ax + by + c Or more commonly, you'll see it used in a simpler form: x y + a x + b y + d xy + ax + by + d But especially when we’re given something like: x y + a x + b y + a b xy + ax + by + ab It becomes very easy to factor. Let me walk you through it step-by-step. 💡 The Key Idea We add and subtract a constant to turn the expression into a perfect rectangle (a.k.a. a factorable quadratic or product of binomials). The “complete the rectangle” version of this trick usually works best on expressions like: x y + a x + b y + a b xy + ax + by + ab We treat it like this: x y + a x + b y + a b = ( x + b ) ( y + a ) xy + ax + by + ab = (x + b)(y + a) ✅ Step-by-Step Example Factor: x y + 3 x + 2 y + 6 xy + 3x + 2y + 6 Step 1: Rearrange the terms: Group like this: x y + 3 x + 2 y + 6 xy + 3x + ...
Read more

IOQM 2023 solutions

Image
Q1. Let n be a positive integer such that 1 ≤ n ≤ 1000 . Let M n be the number of integers in the set X n = { 4 n + 1 , 4 n + 2 , … , 4 n + 1000 } . Let a = max ⁡ { M n : 1 ≤ n ≤ 1000 } , and b = min ⁡ { M n : 1 ≤ n ≤ 1000 } . a = \max\{M_n : 1 \leq n \leq 1000\}, \quad \text{and} \quad b = \min\{M_n : 1 \leq n \leq 1000\}. Find a − b a - b . Solution: Quick tip: If n^2 = k then (n+1)^2 = k + (2n + 1) We will use this here. Also as the numbers grow larger the gap between 2 perfect squares becomes less and less. For. e.g. there are 10 perfect squares between 1 and 100 but only 4 perfect squares between 101 and 200. So X1 = {sqrt(5) ... sqrt(1004)} will have the most number of perfect squares. While X1000 = {sqrt(4001)...sqrt(5000)} will have the least. In X1 the first integer square root is 3 and we know that 1024 is the square of 32 so the last integer will be 31. Total: 31 - 3 + 1 = 29 integers in X1. In X1000, We know that 64^2 = 4096 so 64 is the first integer. 70^2 = 4900 and ...
Read more