IOQM 2022 solutions

 

Now,
BMP ~ BAE. Why? Since Angle B is same in both. BM/BA = BP/BE = 1/2. So SAS similarity criteria.
So: MP/AE = 1/2 => x + z = 2MP => x = MP. But MP = AB/2 = (x+y)/2 => 2x = x + y => x = y.
So ABE is equilateral and hence all angles are 60. So DAE = 60 as well.

Q4.

Solution:
After applying the transformation 4 times you will get:
81M + 80 = 16N + 405
=> 81(M-5) = 16(N-5)
Minimum values of M,N when LHS  = RHS, M = 5, N = 5 and both sides become 0.
Next equal values will be when M-5 = 16 and N-5 = 81 since there are no common factors in 81 and 16.
So M + N = 10


Q5.

Let $a,b\in\mathbb Z_{>0}$ with

$$a^3-b^3-ab=25>0\;\Rightarrow\; a>b.$$

Put $d=a-b\ge1$. Then

$$a^3-b^3-ab=(a-b)(a^2+ab+b^2)-ab = d(3b^2+3bd+d^2)-b(b+d) = d^3+(3d-1)ab.$$

So

$$d^3+(3d-1)ab=25.$$

Now check possible $d$:

• $d\ge3\Rightarrow d^3\ge27>25$, impossible.

• $d=2\Rightarrow 8+5ab=25\Rightarrow ab=17/5$, impossible.

• $d=1\Rightarrow 1+2ab=25\Rightarrow ab=12.$

With $a=b+1$ and $ab=12$, we get $b(b+1)=12\Rightarrow b=3,\ a=4$.

Thus the (unique) solution is $(a,b)=(4,3)$. Hence a^ + b^3 = 43

$\boxed{{\displaystyle \mathrm{}}}$

Solution:
From triangle inequalities:
|x - 18| < y < x + 18
Since x,y are integers => count of possible values of y  = (x + 18) - |x-18| - 1
We know that
(a + b) - |a-b| = 2min(a,b)
Why?
Case 1 : a<= b => a + b -|a-b| = a + b - b + a = 2a = 2.min(a,b)
Case 2: b<=a => a + b - |a-b| = a + b - a + b = 2b = 2.min(a,b)
In both cases it holds, H.P.
So count of possible values of y = 2min(x,18) - 1 = 35
=> min(x,18) = 18
=> x >= 18
But x < 100 => num possible values of x = 82.

Prerequisite: Choose 'k' numbers from 'n' with no two consecutive.

Solution:
Selecting disjoint pairs reduces to selecting non adjacent numbers.
For e.g. from 1,2,3,4 we can pick 3 pairs: 12,23,34 but since they have to be disjoint it means that 12,34 is valid but 12,23 is not.
Given that we have 10 digits, we have 9 pairs to choose from. But none can overlap.
So selecting 'k' disjoint pairs from 10 digits  = selecting 'k' digits from 'n' and none of the digits will be consecutive.
So (9-k+1)Ck
k = 1 => 9C1 (selecting one pair)
k = 2 => 8C2 selecting two pairs and so on..
k = 3 => 7C3
k = 4 => 6C4
k = 5 => 5C5
Total if you add them up  = 88 = answer.

Q11.

Solution:

EPCQ is a cyclic quadrilateral since all points lie on the same circle.
We can apply Ptolemy's theorem.
EP.CQ + PC.QE = EC.PQ
We also know that perpendicular dropped on a chord from the circle center bisects the chord.
CE is a chord and AB is the perpendicular dropped from the center.
So CD = ED = r (where r is the radius of the circle centered at C).
=> CE = 2r
Also CP = CQ = r
=> r (EP + QE) = 2r.PQ
=> (EP + QE) = 2PQ
But PQ + EP + QE = 24
=> 24 - PQ = 2PQ
=> PQ = 8

Solution:

First note how the angles are computed.
B is given as 60 and PQ || BC => QPA = 60.
QRB is 60 because the trapezium is isosceles.
QRC = 180 - 60 = 120
RQC = 180 - 120 - 30 = 30
Since triangle QRC is isosceles => QR = RC.
Let PB = x since PBQR is isosceles trapezium PB = QR = RC = x.
Let BC = L => BR = L - x
From trapezium angles
PQ = BR - BP.cos(60) - QR.(cos60) = BR - 2x.cos(60) = L - x - 2.x.(1/2) = L - 2x
[BPQR] = (BR + PQ)/2 * x.sin(60) = (L - x + L - 2x) * x.sqrt(3)/4 = (2L - 3x) * x.sqrt(3)/4
[ABC] = 2 * L/2 * sqrt(3)L/2 = L^2. sqrt(3)/2
2[ABC]/[BPQR] = 4.sqrt(3).L^2/(2L - 3x) * x.sqrt(3) = 4L^2/(2L - 3x).x
 = [4L^2/x^2] / (2L/x - 3)
Let 2L/x = y, then the expression becomes:
y^2/(2y - 3)
Since all sides have to be positive, we can come  up with some bounds on L,x.
PQ = L - 2x > 0 => L/2 > x =>  L/x = y > 2.
Now coming back to y^2/(2y - 3) = y^2/2.(y-3/2)
Let x = y - 3/2, so it becomes:
(x + 3/2)^2/2x = (x^2 + 9/4 + 3x)/2x = x/2 + 9/8x + 3/2
To find min value of x/2 + 9/8.x, use A.M. G.M. inequality:
(x/2 + 9/8.x) >= 2 sqrt(9/16) = 3/2
So min value of the full expression  = 3/2 + 3/2 = 3

 
Solution 1:
Let angle BAD = w.
Then x + y + z + w = 180 = 3y + w
=> w = 3(60-y)
w > 0 => max value of y is 59 since y is an integer. 
Answer = 59.
But here we didn't use AD = BC.
If we could construct a triangle with y = 59 satisfying AD = BC and A.P. condition, we are done.
It can be shown that y = 59, x = 56, z = 62 gives such a triangle.

Solution 2:

Case 1:
All 3 directions(both ways) are chosen.
Total 6! ways.
But for 'x' direction, this will consider both the cases -> when x1 comes before x2 and when x2 comes before x1. Only half of them are valid. So divide by 2!. Similarly for y and z. Hence 3 times divide by 2!.

Case 2:
2 Directions chosen.
3C1 for choosing first direction.

2C1 for second direction.

6!/2!2! to arrange since x1 and x2 occur twice each.

Now only y1y2 is valid not y2y1 so divide by 2!.

For x1x1x2x2 there are 4!/2!2! = 6 ways to arrange.
But only one is valid: x1x2x1x2 so divide by 6.

Case 3:

Only one direction chosen.
Let's say x.
Only one valid way.
x1x2x1x2x1x2.
Similarly for y,z.
So total 3.

 

Solution:
Each box should leave a different remainder by 6.
Choices are 6n + 1, 6n + 2, 6n + 3, 6n + 4, 6n + 5, 6n + 6
Why did we choose 6n + 6 and not 6n + 0?
So that we are free to put n = 0 and it will help us in stars and bars application which works best when each variable can be 0.
Now let's say we choose remainders 1,2,3,5
So
6n1 + 1 + 6n2 + 2 + 6n3 + 3 + 6n4 + 5 = 52
=> 6.(some integer) = 52 - 11 = not divisible by 6.
So we have to choose in such a way that 52 - (sum of remainders)  mod 6 = 0
=> (sum of remainders) mod 6 = 52 mod 6 = 4
So sum of remainders when divided by 6 should leave the remainder 4.
Maximum possible sum of remainders  = 3 + 4 + 5 + 6 = 18
Minimum = 1+2+3+4 = 10
So we have 10 and 16 as the possible sums of remainders which will leave the remainder 4 when divided by 6.
Let's see how can we make 10:
1+2+3+4 is the only way.
How to make 16:
If we don't take 6 then the largest sum is 2+3+4+5 = 14 so we need 6.
Remaining 10:
2+3+4 still falls short. So we need 5.
Remaining 5:
Only  2 ways: 1 +4 and 2+3
So 
Case 1: (1,2,3,4)
6*(n1+n2+n3+n4) = 52 - 10 = 42 => n1+n2+n3+n4 = 7 => 10C3 using stars and bars
Case 2: (1,4,5,6):
Case 3: (2,3,5,6):
6*(n1+n2+n3+n4) = 52 - 16 = 36 => n1+n2+n3+n4 = 6 => 2*9C3 using stars and bars
Total: 4!(120 + 2*84) = 6912 = 69*100 + 12 => a + b = 69 + 12 = 81
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