IOQM 2023 Apple Mock One

23 June 20237 minutesviews

This article contains spoilers for the mock test.

I recommend attempting the problems before proceeding to read the article. You can download the problems and the answer key using the buttons below.

This is the first ever public IOQM mock test of MOMC. The problems are taken from CHMMC 2012 Spring, PUMaC 2020 and HMMT 2009 Feb Guts Round. All the credit and authorship of the problems belongs to their respective sources.

I had a fun time making the mock, as well as naming it. Last year's mocks' names were of the format <abbreviation of month>_number (Eg: SP3). To switch things up this year, I decided to name them after a fruit instead.

I wasn't able to fit in some nice problems in the test, so I added them as bonus problems at the end. Try them out as well! Now I wish to share with you, the key ideas of some of my favourite problems in the test. So if you are planning to attempt the mock, beware of the spoilers ahead.

Problem 15: (PUMaC 2020 Algebra A P4)\textbf{(PUMaC 2020 Algebra A P4)} Let PP be a 1010-degree monic polynomial with roots r1,r2,,r100r_1, r_2, \ldots, r_{10} \neq 0 and let QQ be a 4545-degree monic polynomial with roots 1ri+1rj1rirj\frac{1}{r_i}+\frac{1}{r_j}-\frac{1}{r_i r_j} where i<ji<j and i,j{1,,10}i, j \in\{1, \ldots, 10\}. If P(0)=Q(1)=2P(0)=Q(1)=2, then log2(P(1))\log _2(|P(1)|) can be written as ab\frac{a}{b} for relatively prime integers a,ba, b. Find a+ba+b.

Solution: This problem is a very cool yet simple example of representing a polynomial as the product of (xαi)(x-\alpha_i) where αi\alpha_i is one of its root. So we write QQ as:

Q(x)=i<j(x1ri1rj+1rirj)Q(x)=\prod_{i<j}\left(x-\frac{1}{r_i}-\frac{1}{r_j}+\frac{1}{r_i r_j}\right)

Now we want to manipulate this expression in order to be able to use the value of P(0)P(0),Q(1)Q(1) and compute P(1)P(1). So we substitute in x=1x = 1 to get: Q(1)=i<j(11ri1rj+1rirj)Q(1)=\prod_{i<j}\left(1-\frac{1}{r_i}-\frac{1}{r_j}+\frac{1}{r_i r_j}\right) Thus we know that this expression is equal to 22. Now note that the denominators resemble something like (1ri)(1rj)(1-r_i)(1-r_j), which reminds us that we can write P(x)P(x) as i=19(1ri)\prod_{i=1}^9 (1-r_i) which is a good sign. Thus taking the common fraction we get:

Q(1)=i<j(rirjrjri+1rirj)=i<j((1ri)(1rj)rirj)=P(1)2(r1r2r10)9Q(1) = \prod_{i<j} \left( \frac{r_ir_j - r_j-r_i+1}{r_ir_j} \right) = \prod_{i<j} \left( \frac{(1-r_i)(1-r_j)}{r_ir_j} \right) = \frac{P(1)^2}{(r_1r_2\cdots r_{10})^9}

Now we can use P(0)=2P(0) = 2 as it is the product of roots, thus the fraction evaluates to P(1)929\frac{P(1)^9}{2^9}. So P(1)9=210P(1)^9 = 2^{10}. From here, the answer is 109\frac{10}{9}.

Onto the next problem!

Problem 24: (CHMMC 2012 Individual P13)\textbf{(CHMMC 2012 Individual P13)} Suppose that a,b,c,d,e,fa, b, c, d, e, f are real numbers such that

a+b+c+d+e+f=0a+2b+3c+4d+2e+2f=0a+3b+6c+9d+4e+6f=0a+4b+10c+16d+8e+24f=0a+5b+15c+25d+16e+120f=42,\begin{gathered} a+b+c+d+e+f=0 \\ a+2 b+3 c+4 d+2 e+2 f=0 \\ a+3 b+6 c+9 d+4 e+6 f=0 \\ a+4 b+10 c+16 d+8 e+24 f=0 \\ a+5 b+15 c+25 d+16 e+120 f=42, \end{gathered}

Find the sum of digits of a+6b+21c+36d+32e+720fa+6 b+21 c+36 d+32 e+720 f

Solution: Even after noticing the pattern in the coefficients, this problem seems very ugly and bashy. But it is actually very nice, believe it or not. This is a beautiful demonstration of finite differences. Firstly define g(k)=a+kb+k(k1)2c+k2d+2ke+k!fg(k) = a + kb + \frac{k(k-1)}{2}c + k^2 d + 2^ke + k!\cdot f. The key observation is that the coefficients of a,b,c,da,b,c,d are polynomials of degree atmost 22. Thus we can feasibly use finite differences. Thus g(k)3g(k1)+3g(k2)g(k3)g(k)-3 g(k-1)+3 g(k-2)-g(k-3) will have a coefficient of 00 for all a,b,c,da,b,c,d. Thus we obtain the following three equations in the three variables e,fe,f and g(6)g(6):

g(4)3g(3)+3g(2)g(1)=0=e+11fg(5)3g(4)+3g(3)g(2)=42=2e+64fg(6)3g(5)+3g(4)g(3)=g(6)126=4e+426f\begin{gathered} g(4)-3 g(3)+3 g(2)-g(1)=0=e+11 f \\ g(5)-3 g(4)+3 g(3)-g(2)=42=2 e+64 f \\ g(6)-3 g(5)+3 g(4)-g(3)=g(6)-126=4 e+426 f \end{gathered}

Solving the first two equations we get f=1f=1 and e=11e = -11. Plugging this into the third equation gives g(6)=508g(6) = 508.

Problem 30: (CHMMC 2012 Individual P11)\textbf{(CHMMC 2012 Individual P11)} Let F0=0,F1=1F_0=0, F_1=1, and Fn=Fn1+Fn2F_n=F_{n-1}+F_{n-2}. How many subsets SS of {1,2,99}\{1,2, \ldots 99\} are there such that

F1001=iSFi?F_{100}-1=\sum_{i \in S} F_i ?

Solution: This problem relies on some basic intuition about size regarding Fibonacci numbers. We claim that, in general for F2n1F_{2n}-1, the answer is nn. We proceed by induction. The base case n=1n = 1 is clear. Now suppose it is true for nn. Notice that F2n+21F_{2n+2}-1 is quite large compared to other fibonacci numbers (especially FiF_i with i2ni \leq 2n). Also remembering the identity k=12nFk=F2n+21\sum_{k=1}^{2n} F_k = F_{2n+2}-1, we are motivated to do casing on whether F2n+1F_{2n+1} is involved in the sum or not. If it is not involved, then the above identity implies that, to get the desired sum, we need to choose all the available fibonacci numbers. Thus there is only 11 way in this case. Now if we do choose F2n+1F_{2n+1} then the rest of the sum is F2n+21F2n+1=F2n1F_{2n+2}-1 - F_{2n+1} = F_{2n} - 1. Now by induction there are nn ways for this. Thus in total we have n+1n+1 ways, completing the inductive step.

Proving the identity is not very difficult, for example, it can be shown by induction. If the result is true for nn then k=12n+2Fk=k=12nFk+F2n+1+F2n+2=F2n+2+F2n+11+F2n+2=F2n+3+F2n+21=F2n+41\sum_{k=1}^{2n+2} F_k = \sum_{k=1}^{2n} F_k + F_{2n+1} + F_{2n+2} = F_{2n+2} + F_{2n+1} - 1 + F_{2n+2} = F_{2n+3} + F_{2n+2} - 1 = F_{2n+4} - 1 completing the induction.

Now we would move on to a funny geometry problem

Problem 23: (CHMMC 2012 Team P10)\textbf{(CHMMC 2012 Team P10)} A convex polygon in the Cartesian plane has all of its vertices on integer coordinates. One of the sides of the polygon is ABA B where A=(0,0)A=(0,0) and B=(51,51)B=(51,51), and the interior angles at AA and BB are both at most 4545 degrees. Assuming no 180180 degree angles, what is the maximum number of vertices this polygon can have?

Solution: As there is almost no condition on the sides of the polygon, we just consider the sides as vectors v1,v2,vkv_1, v_2, \cdots v_k, where kk is the number of sides minus one (Side ABAB isn't considered in these vectors). Observe that the sum of entries of the vectors v1+v2+vkv_1 + v_2 + \cdots v_k is 5151. Clearly all the entries are integers. Infact these integers must be non-negative. This is because the polygon must be convex and that the interior angles at AA and BB are both at most 4545 (Only proceed if you fully understand why). Now to maximize the number of vectors, we need the entries to be as small as possible. We write the "smallest" possible vectors until the sum of the entries is close to or equal to 5151. All the vectors with sum of ii and jj is less than or equal to 77 is:

(1,0),(0,1),(1,1),(2,1),(1,2),(3,1),(1,3),(4,1),(3,2),(2,3),(1,4),(5,1),(1,5),(6,1),(5,2),(4,3),(3,4),(2,5),(1,6),\begin{gathered} (1,0),(0,1),(1,1),(2,1),(1,2),(3,1),(1,3),(4,1),(3,2), \\ (2,3),(1,4),(5,1),(1,5),(6,1),(5,2),(4,3),(3,4),(2,5),(1,6), \end{gathered}

The sum of the entries is 4646. Now we show that k20k \leq 20. Suppose for the sake of contradiction that k21k \geq 21. Then we need to add atleast 22 vectors with sum of both entries atleast 88. Thus sum of all entries is atleast 462+82=108>10246 \cdot 2 + 8 \cdot 2 = 108 > 102.

The following construction is due to CHMMC Official Solutions: Take the list of 1919 vectors above along with (3,5)(3,5), and order them by the angle they make with the vector (1,0)(1,0) from smallest to largest (so the list starts with (1,0)(1,0) and ends with (0,1)(0,1) ). Make the sides of the polygon correspond to the vectors in this order and then replace (1,0)(1,0) with (3,0)(3,0). The increasing angle condition makes the polygon convex, the interior angles at AA and BB are exactly 45 degrees, and the sum of all the vectors is (51,51)(51,51).