# Pre-RMO training; a statement and its converse; logic and plane geometry

I hope the following explanation is illuminating to my readers/students:

How to prove that two lines are parallel ? (Note that we talk of parallel lines only when they lie in the same plane; on the other hand: consider the following scenario — your study table and the floor on which it stands. Let us say you draw a straight line AB on your study table and another line PQ on the floor on which the study table is standing; then, even though lines AB and PQ never meet, we do not say that they are parallel because they lie in different planes. Such lines are called skew lines. They are dealt with in solid geometry or 3D geometry or vector spaces).

Coming back to the question — when can we say that two lines are parallel?

Suppose that a transversal crosses two other lines.

1) If the corresponding angles are equal, then the lines are parallel.
2) If the alternate angles are equal, then the lines are parallel.
3) If the co-interior angles are supplementary, then the lines are parallel.

A STATEMENT AND ITS CONVERSE

Let us first consider the following statements:

A transversal is a line that crosses two other lines. If the lines crossed by a transversal are parallel, then the corresponding angles are equal; if the lines crossed by a transversal are parallel, then the alternate interior angles are equal; if the lines crossed by a transversal are parallel, then the co-interior angles are supplementary.

The statements given below are the converses of the statement given in the above paragraph; meaning that they are formed from the former statements by reversing the logic. For example:

STATEMENT: If the lines are parallel then the corresponding angles are equal.

CONVERSE: If the corresponding angles are equal, then the lines are parallel.

Pairs such as these, a statement and its converse, occur routinely through out mathematics, and are particularly prominent in geometry. In this case, both the statement and its converse are true. It is important to realize that a statement and its converse are, in general, quite different. NEVER ASSUME THAT BECAUSE A STATEMENT IS TRUE, SO ITS CONVERSE IS ALSO TRUE. For example, consider the following:

STATEMENT: If a number is a multiple of 4, then it is even.
CONVERSE: If a number is even, then it is a multiple of 4.

The first statement is clearly true. But, let us consider the number 18. It is even. But 18 is not a multiple of 4. So, the converse is not true always.

$\it Here \hspace{0.1in} is \hspace{0.1in}an \hspace {0.1in}example \hspace{0.1in}from \hspace{0.1in}surfing$

STATEMENT: If you catch a wave, then you will be happy.
CONVERSE: If you are happy, then you will catch a wave.

Many people would agree with the first statement, but everyone knows that its converse is plain silly — you need skill to catch waves.

Thus, the truth of a statement has little to do with its converse. Separate justifications (proofs) are required for the converse and its statements.

Regards,
Nalin Pithwa.

Reference: (I found the above beautiful, simple, lucid explanation in the following text): ICE-EM, year 7, book 1; The University of Melbourne, Australian Curriculum, Garth Gaudry et al.

# Pre RMO or RMO Solutions: Homi Bhabha Science Foundation

Problem:

Reference: Problem Primer for the Olympiad by C. R. Pranesachar, et al. Prism Books

Five men A, B, C, D, E are wearing caps of black or white colour without each knowing the colour of his own cap. It is known that a man wearing a black cap always speaks the truth while a man wearing a white cap always lies. If they make the following statements, find the colour of the cap worn by each of them:

A: I see three black and one white cap.

B: I see four white caps.

C: I see one black and three white caps.

D: I see four black caps.

Solution:

Suppose E is wearing a white cap:

Then, D is lying and hence must be wearing a white cap. Since D and E both have white caps, A is lying and hence, he must be wearing white cap. If C is speaking truth, then C must be wearing a black cap and B must be wearing a black cap as observed by C. But, then B must observe a black cap on C. Hence, B must be lying. This implies that B is wearing a white cap which is a contradiction to C’s statement.

On the other hand, if C is lying, then C must be wearing a white cap. Thus, A, C, D and E are wearing white caps which makes B’s statement true. But, then B must be wearing a black cap and this makes C statement correct.

Thus, E must be wearing a black cap. This implies that B is lying and hence, must be having a white cap. But, then D is lying and hence, must be having a white cap since B and D have white caps. A is not saying the truth. Hence, A must be wearing a white cap. These together imply that C is truthful. Hence, C must be wearing a black cap. Thus, we have the following distribution:

A: white cap; B: white cap; C: black cap; D: white cap; E: black cap.

Hope you enjoyed it! There can be some other approaches too starting with some other assumption(s).

Nalin Pithwa.

# Pre-RMO or RMO sample problem solutions: logic questions

Reference: Problem Primer for the Olympiad by C. R. Pranesachar et al, Prism Books.

Problem 1:

The sixty four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the integers on the neighbouring squares. (Two squares are neighbours if they share a common edge or vertex. Thus, a square can have 8, 5, or 3 neighbours depending on its position). Show that all the sixty four entries are in fact equal.

Solution 1:

Consider the smallest value among the 64 entries on the board. Since it is the average of the surrounding numbers, all those numbers must be equal to this number as it is the smallest. This gives some more with the smallest value. Continue in this way till all the squares are covered.

Problem 2:

Let T be the set of all triples $(a,b,c)$ of integers such that $1 \leq a < b . For each triple $(a,b,c)$ in F, take the product abc. Add all these products corresponding to all triples in T. Prove that the sum is divisible by 7.

Solution 2:

For every triplet $(a,b,c)$ the triplet $(7-c,7-b,7-a)$ is in T and these two are distinct as $7 \neq 2b$. Pairing off $(a,b,c)$ with $(7-c,7-b,7-a)$ for each $(a,b,c) \in T$. 7 divides $abc-(7-c)(7-b)(7-a)$.

Problem 3:

In a class of 25 students, there are 17 cyclists, 13 swimmers, and 8 weight lifters and no one is all the three. In a certain mathematics examination, 6 students got grades D or E. If the cyclists, swimmers and weight lifters all got grade B or C, determine the number of students who got grade A. Also, find the number of cyclists who are swimmers.

Solution 3:

Let S denote the set of all 25 students in the class, X the set of swimmers in S, Y the set of all weight lifters and Z the set of all cyclists. Since students in $X \bigcup Y \bigcup Z$ all get grades B and C and six students get grades D or E, the number of students in $X \bigcup Y \bigcup Z \leq 25-6=19$. Now assign one point to each of the 17 cyclists, 13 swimmers, and 8 weight lifters. Thus, a total of 38 points would be assigned among the students in $X \bigcup Y \bigcup Z$. Note that no student can have more than two points as no is all the three. Then, we should have $|X\bigcup Y \bigcup Z| \geq 19$ as otherwise 38 points cannot be accounted for. (For example, if there were only 18 students in $X \bigcup Y \bigcup Z$ the maximum number of points that could be assigned to them is 36). Therefore, $|X \bigcup Y \bigcup Z|=19$ and each student in  $X \bigcup Y \bigcup Z$ is in exactly 2 of the sets X, Y and Z. Hence, the number of students getting grade $A=25 - 19-6=0$, that is, no student gets A grade. Since there are $19-8=11$ students who are not weight lifters all these 11 students must be both swimmers and cyclists. (Similarly, there are 2 who are both swimmers and weight lifters and 6 who are both cyclists and weight lifters).

More later,

Nalin Pithwa

# Pre-RMO or RMO tutorial: Homi Bhabha Science Foundation

Some more questions based on logic only:

1. Five men A, B, C, D, E are wearing caps of black or white colour without each knowing the colour of his cap. It is known that a man wearing a black cap always speaks the truth while a man wearing a white cap always lies. If they made the following statements, find the colour of the cap worn by each of them:

A: I see three black and one white cap.

B: I see four white caps.

C: I see one black and three white caps.

D: I see four black caps.

Nalin Pithwa.

# RMO or Pre-RMO training: Homibhabha Science Foundation exam

Sample questions based on logic only:

1. Given the sixty-four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the integers on the neighbouring squares. (Two squares are neighbours if they share a common edge or vertex. Thus, a square can have 8, 5 or 3 neighbours depending on its position). Show that alll the sixty four entries are in fact equal.
2. Let T be the set of all triples $(a,b,c)$ of integers such that $1 \leq a . For each triple $(a,b,c)$ in T, take the product abc. Add all these products corresponding to all triples in F. Prove that the sum is divisible by 7.
3. In a class of 25 students, there are 17 cyclists, 13 swimmers and 8 weight lifters and no one is all the three. In a certain mathematics examination, 6 students get grades D or E. If the cyclists, swimmers and weight lifters all got grade B or C determine the number of students, who got grade A. Also, find the number of cyclists who are swimmers.

Any ideas ? Please wait for solutions tomorrow.

Nalin Pithwa.