RMO Geometry DPP 4


 



Solution:
Part 1:
(AEHZ) is cyclic quadrilateral since opposite angles HZA and HEA sum to 180.
Its diameter is AH since it subtends 90 deg at circumference.
N is the center since it's the midpoint of AH.
 
(NZMDE) are also concyclic since all these fall on 9-point circle of ABC.
EZ is the common chord of these 2 circles and hence the radical axis.
Angle NDM = 90 => MN is a diameter of this 9-point circle.
Hence the center of this will lie on MN.
So both the centers lie on the line MN.
Line joining centers is perpendicular to the radical axis. H.P.

Part 2:
MN is perpendicular to EZ as both the centers lie on MN.
The line containing both the centers will bisect the chord EZ in circle (AEHZ).
In triangle AEZ, MN is the perpendicular bisector of EZ.
So using Arc midpoint lemma MN and AI will meet on the circumcircle of AEZ and that point is K.
Using the same lemma again MN and AI' will meet on the circumcircle again at L.
K,N,L are colinear hence KL is a diameter.
Hence KL = AH since AH is also a diameter of the same circle.



Additionally, Q is the midpoint of EC.





Proof of part 1: M,L,S line on 9-point circle of triangle EFC. The same circle will also pass through midpoint of EC.

Part 2: ED || QN by midpoint theorem in triangle EDC. N is midpoint of DC and Q is midpoint of EC.

In triangle FDC, using midpoint theorem, NS || FD => NS || DB since F,D,B are colinear.

Angle EDB has 2 rays ED and DB.
Angle QNS has 2 rays QN and NS.
Since the corresponding rays are parallel, the angles formed by them are equal.

Part 3:
Circumcircle of MLS is the 9-point circle of triangle EFC and hence it will also pass through Q which is EC's midpoint.
If we prove that QLSN is a cyclic quadrilateral then N will also lie on that same circle since a circle passing through 3 non-colinear points is unique.

In part 2 we proved that Angle QNS = Angle EDB
Angle EDB = Angle ADB (E,A,D are colinear)
Angle ADB = Angle ACB (A,B,C,D are concyclic, same chord, same angles)
Angle ACB = Angle SCB (A,S,C are colinear)
Angle SCB  = Angle SCL (C,L,B are colinear)
In right angle triangle FLC, S is the midpoint of hypotenuse FC hence SC = SF = SL
=> SLC is an isosceles triangle => Angle SLC = Angle SCL
=> Angle QNS = Angle SLC
But in the quadrilateral QLSN, this means that external angle SLC = opposite internal angle QNS
=> QLSN is a cyclic quadrilateral.
=> N will also line on (MLSQ)
Similarly N will also line on (MTKP) where P is the midpoint of ED.
Hence proved.

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

Combinatorics DPP - RACE 6 - Q16 pending discussion

Image
Q1 . A question paper consists of two sections. Section A has 7 questions and section B has 8 questions. A student has to answer 10 questions out of these 15 questions. The number of ways in which he can answer if he must answer at least 3 of section A and at least 4 of the section B is: Solution: One wrong way to solve this: For the mandatory questions: 7C3*8C4 Now 3 more questions need to be answered from 4 of A and 4 of B. Cases: 3A,0B | 2A,1B | 1A,2B | 0A,3B 4C3*4C0 + 4C2*4C1 + 4C1*4C2 + 4C0*4C1 = 4 + 24 + 24 + 4 = 56 Finally: 7C3*8C4*56 = 35*70*56 = 137200 Why is it wrong? Let's say we chose A1,A2,A3 as mandatory from A and then chose A4,A5,A6 when we did 4C3*4C0 for (3A,0B). But we could have done it the other way round also, i.e. choose A4,A5,A6 as mandatory and then choose A1,A2,A3. So there is overcounting. Correct Solution : Simply create the cases for A,B: (3,7),(4,6),(5,5),(6,4) 7C3*8C7 + 7C4*8C6 + 7C5*8C5 + 7C6*8C4 = 2926. Q2 . A kindergarten teacher has 25 kids in her...
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