A primer for preRMO and RMO plane geometry with basic exercises

Plane geometry is axiomatic deductive logic. I present a quick mention/review of “proofs” which can be “derived” in sequence….building up the elementary theorems …so for example, if there is a question like: prove that the three medians of a triangle are concurrent, please do not use black magic complicated machinery like Ceva’s theorem,etc; or even if say, the question asks you to prove Ceva’s theorem only, you have to prove it using elementary theorems like the ones presented below:

For the present purposes, I am skipping axioms and basic definitions and hypothetical constructions. I am using straight away the reference (v v v old text) : A School Geometry, Metric Edition by Hall and Stevens. (available almost everywhere in India):

Theorem 1:

The adjacent angles which one straight line makes with another straight line on one side of it are together equal to two right angles.

Corollary 1 of Theorem 1:

if two straight lines cut another, the four angles so formed are together equal to four right angles.

Corollary 2 of Theorem 1:

When any number of straight lines meet at a point, the sum of the consecutive angles so formed is equal to four right angles.

Corollary 3 of Theorem 1:

(a) Supplements of the same angle are equal. (ii) Complements of the same angle are equal.

Theorem 2 (converse of theorem 1):

If, at a point in a straight line, two other straight lines, on opposite sides of it, make the adjacent angles together equal to two right angles, then these two straight lines are in one and the same straight line.

Remark: this theorem can be used to prove stuff like three points are in a straight line.

Theorem 3:

If two straight lines cut one another, the vertically opposite angles are equal.

Theorem 4: SAS Test of Congruence of Two Triangles:

If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, then the triangles are equal in all respects.

Theorem 5:

The angles at the base of an isosceles triangle are equal.

Corollary 1 of Theorem 5:

If the equal sides AB, AC of an isosceles triangle are produced, the exterior angles EBC, FCB are equal, for they are the supplements of the equal angles at the base.

Corollary 2 of Theorem 5:

If a triangle is equilateral, it is also equiangular.

Theorem 6:

If two angles of a triangle are equal to one another, then the sides which are opposite to the equal angles are equal to one another.

Corollary of Theorem 6:

Hence, if a triangle is equiangular, it is also equilateral.

Theorem 7 (SSS Test of Congruence of Two Triangles):

If two triangles have the three sides of the one equal to the three sides of the others, each to each, they are equal in all respects.

Theorem 8:

If one side of a triangle is produced then the exterior angle is greater than either of the interior opposite angles.

Corollary 1 to Theorem 8:

Any two angles of a triangle are together less than two right angles.

Corollary 2 to Theorem 8:

Every triangle must have at least two acute angles.

Corollary 3 to Theorem 8:

Only one perpendicular can be drawn to a straight line from a given point outside it.

Theorem 9 :

If one side of a triangle is greater than another, then the angle opposite of the greater side is greater than the angle opposite to the less.

Theorem 10:

If one angle of a triangle is greater than another, then the side opposite to the greater angle is greater than the side opposite to the less.

Theorem 11: Triangle Inequality:

Any two sides of a triangle are together greater than the third side.

Theorem 12: Another inequality sort of theorem:

Of all straight lines drawn from a given point to a given straight line the perpendicular is the least.

Corollary 1 to Theorem 12:

Hence, conversely, since there can be only one perpendicular and one shortest line from O to AB: if OC is the shortest straight line from O to AB, then OC is perpendicular to AB.

Corollary 2 to Theorem 12:

Two obliques OP, OQ which cut AB at equal distance from C, the foot of the perpendicular are equal.

Corollary 3 to Theorem 12:

Of two obliques OQ, OR, if OR cuts AB at the greater distance from C. the foot of the perpendicular, then OR is greater than OQ.

Theorem 13 :

If a straight line cuts two other straight lines so as to make: (i) the alternate angles equal or (ii) an exterior angle equal to the interior opposite angle on the same side of the cutting line or (iii) the interior angles on the same side equal to two right angles, then in each case, the two straight lines are parallel.

Theorem 14:

If a straight line cuts two parallel lines, it makes : (i) the alternate angles equal to one another; (ii) the exterior angle equal to the interior opposite angle on the same side of the cutting line (iii) the two interior angles on the same side together equal to two right angles.

Theorem 15:

Straight lines which are parallel to the same straight line are parallel to one another.

Theorem 16:

Sum of three interior angles of a triangle is 180 degrees.

Also, if a side of a triangle is produced the exterior angle is equal to the sum of the two interior opposite angles.

Corollary 1:

All the interior angles of one rectilinear figure, together with four right angles are equal to twice as many right angles as the figure has sides.

Corollary 2:

If the sides of a rectilinear figure, which has no reflex angle, are produced in order, then all the exterior angles so formed are together equal to four right angles.

Theorem 17: AAS test of congruence of two triangles:

If two triangles have two angles of one equal to two angles of the other, each to each, and any side of the first equal to the corresponding side of the other, the triangles are equal in all respects.

Theorem 18:

Two right angled triangles which have their hypotenuses equal, and one side of one equal to one side of the other are equal in all respects.

Theorem 19:

If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle included by the two sides of one greater than the angle included by the two corresponding sides of the other, then the base of that which has the greater angle is greater than the base of the other.

Conversely,

if two triangles have two sides of the one equal to two sides of the other, each to each, but the base of one greater than the base of the other, then the angle contained by the sides of that which has the greater base is greater than the angle contained by the corresponding sides of the other.

Theorem 20:

The straight lines which join the extremities of two equal and parallel straight lines towards the same parts are themselves equal and parallel.

Theorem 21:

The opposite sides and angles of a parallelogram are equal to one another, and each diagonal bisects the parallelogram.

Corollary 1 to Theorem 21:

If one angle of a parallelogram to a right angle, all its angles are equal.

Corollary 2 to Theorem 21:

All the sides of a square are equal and all its angles are right angles.

Corollary 3 to Theorem 21:

The diagonals of a parallelogram bisect each other.

Theorem 22:

If there are three or more parallel straight lines, and the intercepts made by them on any transversal are equal, then the corresponding intercepts on any other transversal are also equal.

Tutorial exercises based on the above:

Problem 1: In the triangle ABC, the angles ABC, ACB are given equal. If the side BC is produced both ways, show that the exterior angles so formed are equal.

Problem 2: In the triangle ABC, the angles ABC, ACB are given equal. If AB and AC are produced beyond the base, show that the exterior angles so formed are equal.

Problem 3: Prove that the bisectors of the adjacent angles which one straight line makes with another contain a right angle. That is to say, the internal and external bisectors of an angle are at right angles to one another.

Problem 4: If from O a point in AB two straight lines OC, OD are drawn on opposite sides of AB so as to make the angle COB equal to the angle AOD, show that OC and OD are in the same straight line.

Problem 5: Two straight lines AB, CD cross at O. If OX is the bisector of the angle BOD, prove that XO produced bisects the angle AOC.

Problem 6: Two straight lines AB, CD cross at O. If the angle BOD is bisected by OX, and AOC by OY, prove that OX, OY are in the same straight line.

Problem 7: Show that the bisector of the vertical angle of an isosceles triangle (i) bisects the base (ii) is perpendicular to the base.

Problem 8: Let O be the middle point of a straight line AB, and let OC be perpendicular to it. Then, if P is any point in OC, prove that PA=PB.

Problem 9: Assuming that the four sides of a square are equal, and that its angles are all right angles, prove that in the square ABCD, the diagonals AC, BD are equal.

Problem 10: Let ABC be an isosceles triangle: from the equal sides AB, AC two equal parts AX, AY are cut off, and BY and CX are joined. Prove that BY=CX.

Problem 11: ABCD is a four-sided figure whose sides are all equal, and the diagonal BD is drawn : show that (i) the angle ABD = the angle ADB (ii) the angle CBD = the angle CDB (iii) the angle ABC = the angle ADC.

Problem 12: ABC, DBC are two isosceles triangles drawn on the same base BC, but on opposite sides of it: prove that the angle ABD = the angle ACD.

Problem 13: ABC, DBC are two isosceles triangles drawn on the same base BC, but on the same side of it: prove that the angle ABD = the angle ACD.

Problem 14: AB, AC are the equal sides of an isosceles triangle ABC, and L, M, N are the middle points of AB, BC and CA respectively; prove that (i) LM = NM (ii) BN = CL (iii) the angle ALM = the angle ANM.

Problem 15: Show that the straight line which joins the vertex of an isosceles triangle to the middle points of the base (i) bisects the vertical angle (ii) is perpendicular to the base.

Problem 16: If ABCD is a rhombus, that is, an equilateral four sided figure, show by drawing the diagonal AC that (i) the angle ABC = the angle ADC (ii) AC bisects each of the angles BAD and BCD.

Problem 17: If in a quadrilateral ABCD the opposite sides are equal, namely, AB = CD and AD=CB, prove that the angle ADC = the angle ABC.

Problem 18: If ABC and DBC are two isosceles triangles drawn on the same base BC, prove that the angle ABD = the angle ACD, taking (i) the case where the triangles are on the same side of BC (ii) the case where they are on the opposite sides of BC.

Problem 19: If ABC, DBC are two isosceles triangles drawn on opposite sides of the same base BC, and if AD be joined, prove that each of the angles BAC, BDC will be divided into two equal parts.

Problem 20: Show that the straight lines which join the extremities of the base of an isosceles triangle to the middle points of the opposite sides are equal to one another.

Problem 21: Two given points in the base of an isosceles triangle are equidistant from the extremities of the base: show that they are also equidistant from the vertex.

Problem 22: Show that the triangle formed by joining the middle points of the sides of an equilateral triangle is also equilateral.

Problem 23: ABC is an isosceles triangle having AB equal to AC, and the angles at B and C are bisected by BC and CO: prove that (i) BO = CO (ii) AO bisects the angle BAC.

Problem 24: Show that the diagonals of a rhombus bisect one another at right angles.

Problem 25: The equal sides BA, CA of an isosceles triangle BAC are produced beyond the vertex A to the points E and F, so that AE is equal to AF and FB, EC are joined: prove that FB is equal to EC.

Problem 26: ABC is a triangle and D any point within it. If BD and CD are joined, the angle BDC is greater than the angle BAC. Prove this (i) by producing BD to meet AC (ii) by joining AD, and producing it towards the base.

Problem 27: If any side of a triangle is produced both ways, the exterior angles so formed are together greater than two right angles.

Problem 28: To a given straight line, there cannot be drawn from a point outside it more than two straight lines of the same given length.

Problem 29: If the equal sides of an isosceles triangle are produced, the exterior angles must be obtuse.

Note: The problems 30 to 43 are based on triangle inequalities:

Problem 30: The hypotenuse is the greatest side of a right angled triangle.

Problem 31: The greatest side of any triangle makes acute angles with each of the other sides.

Problem 32: If from the ends of a side of a triangle, two straight lines are drawn to a point within the triangle, then those straight lines are together less than the other two sides of the triangle.

Problem 33: BC, the base of an isosceles triangle ABC is produced to any point D; prove that AD is greater than either of the equal sides.

Problem 34: If in a quadrilateral the greatest and least sides are opposite to one another, then each of the angles adjacent to the least side is greater than its opposite angle.

Problem 35: In a triangle, in which OB, OC bisect the angles ABC, ACB respectively: prove that if AB is greater than AC, then OB is greater than OC.

Problem 36: The difference of any two sides of a triangle is less than the third side.

Problem 37: The sum of the distances of any point from the three angular points of a triangle is greater than half its perimeter.

Problem 38: The perimeter of a quadrilateral is greater than the sum of its diagonals.

Problem 39: ABC is a triangle, and the vertical angle BAC is bisected by a line which meets BC in X, show that BA is greater than BX, and CA greater than CX. Obtain a proof of the following theorem : Any two sides of a triangle are together greater than the third side.

Problem 40: The sum of the distance of any point within a triangle from its angular points is less than the perimeter of the triangle.

Problem 41: The sum of the diagonals of a quadrilateral is less than the sum of the four straight lines drawn from the angular points to any given point. Prove this, and point out the exceptional case.

Problem 42: In a triangle any two sides are together greater than twice the median which bisects the remaining side.

Problem 43: In any triangle, the sum of the medians is less than the perimeter.

Problem 44: Straight lines which are perpendicular to the same straight line are parallel to one another.

Problem 45: If a straight line meets two or more parallel straight lines, and is perpendicular to one of them, it is also perpendicular to all the others.

Problem 46: Angles of which the arms are parallel each to each are either equal or supplementary.

Problem 47: Two straight lines AB, CD bisect one another at O. Show that the straight line joining AC and BD are parallel.

Problem 48: Any straight line drawn parallel to the base of an isosceles trianlge makes equal angles with the sides.

More later. Get cracking. This perhaps the simplest introduction, step by step, to axiomatic deductive logic…discovered by Euclid about 2500 years before ! Hail Euclid !

Cheers,

Nalin Pithwa

Solutions to “next number in sequence”: preRMO, pRMO and RMO

What is the next number in sequence?

A) 15, 20, 20, 6, 6, 19, 19, 5, 14, 20, 5, ?

Solution to A:

Ans is 20. The sequence is the position in the letter of the alphabet of the first letter in the numbers 1 to 12, when given in full. e.g. ONE: O=15.

B) 1, 8, 11, 18, 80, ?

Ans is 81. The sequence comprises whole numbers beginning with a vowel.

C) 1, 2, 4, 14, 21, 22, 24, 31, ?

Ans is 32, The sequence comprises whole numbers containing the letter O.

D) 4, 1, 3, 1, 2, 4, 3, ?

Ans. is 2. The sequence is as follows: there is one number between the two I’s, two numbers between the two 2’s, three numbers between the two 3’s and four numbers between the two 4’s.

E) 1, 2, 4, 7, 28, 33, 198, ?

Ans is 205. $1 + 1 \times 2 + 3 \times 4 + 5 \times 6 + 7$

F) 17, 8, 16, 23, 28, 38, 49, 62, ?

Answer is 70. Sum of digits in all previous numbers in the sequence.

G) 27, 216, 279, 300, ?

Ans is 307. Difference divided by 3 and added to the last number.

H) 9,7,17,79,545, ?

Answer is 4895. Each number is multiplied by its rank in the sequence, and the next number is subtracted.

$9 \times 1 - 2 = 7 \times 3 -4 = 17 \times 5 - 6 = 79 \times 7 - 8 = 545 \times 9 - 10 = 4895$

I) 2,3,10,12,13, 20,?

Answer is 21. They all begin with the letter T.

J) 34, 58, 56, 60, 42, ?

Answer is 52. The numbers are the totals of the letters in the words ONE, TWO, THREE, FOUR, FIVE, SIX when A=1, B=2, C=3, etc.

Regards,

Nalin Pithwa.

Next number in sequence: PreRMO, pRMO, RMO

What is the next number in the sequence?

a) 15, 20, 20, 6, 6, 19, 19, 5, 14, 20, 5, ?

b) 1,8,11, 18, 80, ?

c) 1,2,4,14,21,22,24,31,?

d) 4,1,3,1,2,4,3,?

e) 1,2,4,7,28,33,198,?

f) 17,8,16,23, 28, 38, 49, 62, ?

g) 27, 216, 279, 300, ?

h) 9,7,17,79,545,?

i) 2,3,10,12,13,20,?

j) 34, 58, 56, 60, 42, ?

Regards,

Nalin Pithwa.

Miscellaneous Questions: part I: solution to chess problem by my student RI

Some blogs away I had posted several interesting, non-trivial, yet do-able-with-some-effort problems for preRMO and RMO.

A student of mine, RI has submitted the following beautiful solution to the chess problem. I am reproducing the question for convenience of the readers:

Question:

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 on 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.

Let us denote the set of all integers on the chess board by S (assume they are distinct). [Now, we can use the Well-ordering principle: every non-empty set of non-negative integers contains a least element. That is, every non-empty set S of non-negative integers contains an element a in S such that $a \leq b$ for all elements b of S}. So, also let “a” be the least element of set S here. As it is the average of the neighbouring elements, it can’t be less than each of them. But it can’t be greater than all of them also. So, all the elements of S are equal.

QED.

Three cheers for RI 🙂 🙂 🙂

Regards,

Nalin Pithwa

Miscellaneous questions: part II: solutions to tutorial practice for preRMO and RMO

Refer the blog questions a few days before:

Question 1:

Let $a_{1}, a_{2}, \ldots, a_{10}$ be ten real numbers such that each is greater than 1 and less than 55. Prove that there are three among the given numbers which form the lengths of the sides of a triangle.

Without loss of generality, we may take $1…..call this relation (i).

Let, if possible, no three of the given numbers be the lengths of the sides of a triangle. (That is, no three satisfy the triangle inequality. Note that when we say three numbers a, b and c satisfy the triangle inequality —- it means all the following three inequalities have to hold simultaneously: $a+b>c$, $a+c>b$ and $b+c>a$). We will consider triplets $a_{i}, a_{i+1}, a_{i+2}$ and $1 \leq i \leq 8$. As these numbers do not form the lengths of the sides of a triangle, the sum of the smallest two numbers should not exceed the largest number, that is, $a_{i}+a_{i+1} \leq a_{i+2}$. Hence, we get the following set of inequalities:

$i=1$ gives $a_{1}+a_{2} \leq a_{3}$ giving $2 < a_{3}$.

$i=2$ gives $a_{2}+a_{3} \leq a_{4}$ giving $3 < a_{4}$

$i=3$ gives $a_{3}+a_{4} \leq a_{5}$ giving $5 < a_{5}$

$i=4$ gives $a_{4}+a_{5} \leq a_{6}$ giving $8 < a_{6}$

$i=5$ gives $a_{5}+a_{6} \leq a_{7}$ giving $13 < a_{7}$

$i=6$ gives $a_{6}+a_{7} \leq a_{8}$ giving $21 < a_{8}$

$i=7$ gives $a_{7}+a_{8} \leq a_{9}$ giving $34 < a_{9}$

$i=8$ gives $a_{8}+a_{9} \leq a_{10}$ giving $55

contradicting the basic hypothesis. Hence, there exists three numbers among the given numbers which form the lengths of the sides of a triangle.

Question 2:

In a collection of 1234 persons, any two persons are mutual friends or enemies. Each person has at most 3 enemies. Prove that it is possible to divide the collection into two parts such that each person has at most 1 enemy in his sub-collection.

Let C denote the collection of given 1234 persons. Let $\{ C_{1}, C_{2}\}$ be a partition of C. Let $e(C_{1})$ denote the total number of enemy pairs in $C_{1}$. Let $e(C_{2})$ denote the total number of enemy pairs in $C_{2}$.

Let $e(C_{1}, C_{2})= e(C_{1})+e(C_{2})$ denote the total number of enemy pairs corresponding to the partition $\{ C_{1}, C_{2}\}$ of C. Note $e(C_{1}, C_{2})$ is an integer greater than or equal to zero. Hence, by Well-Ordering Principle, there exists a partition having the least value of $e(C_{1}, C_{2})$.

Claim: This is “the” required partition.

Proof: If not, without loss of generality, suppose there is a person P in $C_{1}$ having at least 2 enemies in $C_{1}$. Construct a new partition $\{D_{1}, D_{2}\}$ of C as follows: $D_{1}=C_{1}-\{ P \}$ and $D_{2}=C_{2}- \{P\}$. Now, $e(D_{1}, D_{2})=e(D_{1})+e(D_{2}) \leq \{ e(C_{1})-2\} + \{ e(C_{2})+1\}=e(C_{1}, C_{2})-1$. Hence, $e(D_{1}, D_{2}) contradicting the minimality of $e(C_{1}, C_{2})$. QED.

Problem 3:

A barrel contains 2n balls, numbered 1 to 2n. Choose three balls at random, one after the other, and with the balls replaced after each draw.

What is the probability that the three element sequence obtained has the properties that the smallest element is odd and that only the smallest element, if any is repeated?

The total number of possible outcomes is $N=2n \times 2n \times 2n=8n^{3}$. To find the total number of favourable outcomes we proceed as follows:

Let a be any odd integer such that $1 \leq a \leq 2n-1$ and let us count the sequences having a as least element.

(i) There is only one sequence $(a,a,a)$ with a repeated thrice.

(ii) There are $2n-a$ sequences of the form $(a,a,b)$ with $a. For each such sequence there are three distinct permutations possible. Hence, there are in all $3(2n-a)$ sequences with a repeated twice.

iii) When $n>1$, for values of a satisfying $1 \leq a \leq (2n-3)$, sequences of the form $(a,b,c,)$ with $a are possible and the number of such sequences is $r=1+2+3+\ldots+(2n+a-1)=\frac{1}{2}(2n-a)(2n-a-1)$. For each such sequence, there are six distinct permutations possible. Hence, there are $6r=3(2n-a)(2n-a-1)$ sequences in this case.

Hence, for odd values of a between 1 and $2n-1$, the total counts of possibilities $S_{1}$, $S_{2}$, $S_{3}$ in the above cases are respectively.

$S_{1}=1+1+1+\ldots+1=n$

$S_{2}=3(1+3+5+\ldots+(2n-1))=3n^{2}$

$3(2 \times 3 + 4 \times 5 + \ldots+ (2n-2)(2n-1))=n(n-1)(4n+1)$.

Hence, the total number A of favourable outcomes is $A=S_{1}+S_{2}+S_{3}=n+3n^{2}+n(n-1)(4n+1)=4n^{3}$. Hence, the required probability is $\frac{A}{N} = \frac{4n^{3}}{8n^{3}} = \frac{1}{2}$. QED>

Cheers,

Nalin Pithwa

Miscellaneous questions: part I : solutions: tutorial practice preRMO and RMO

The following questions were presented in an earlier blog (the questions are reproduced here) along with solutions. Please compare your attempts/partial attempts too are to be compared…that is the way to learn:

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 in 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 squares 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 squares 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 $1 \leq a < b < c \leq 6$. For each triple $(a,b,c)$ in T, 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 triple $(a,b,c)$ in T, the triple $(7-c,7-b,7-a)$ 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 $abc-(7-a)(7-b)(7-c)$.

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 math 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 weight lifters in S, 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 2 points as no one is all three (swimmer, cyclist and weight lifter). 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 would be 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.)

Problem 4:

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 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 4:

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 having a black cap and B must be wearing a black cap as observed by C. But then B must observe a 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 also 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: A: white cap; B: white cap; C:black cap; D:white cap; E: black cap.

Problem 5:

Let f be a bijective function from the set $A=\{ 1,2,3,\ldots,n\}$ to itself. Show that there is a positive integer $M>1$ such that $f^{M}(i)=f(i)$ for each $i \in A$. Here $f^{M}$ denotes the composite function $f \circ f \circ \ldots \circ f$ repeated M times.

Solution 5:

Let us recall the following properties of a bijective function:

a) If $f:A \rightarrow A$ is a bijective function, then there is a unique bijective function $g: A \rightarrow A$ such that $f \circ g = g \circ f=I_{A}$ the identity function on A. The function g is called the inverse of f and is denoted by $f^{-1}$. Thus, $f \circ f^{-1}=I_{A}=f^{-1}\circ f$

b) $f \circ I_{A} = f = I_{A} \circ f$

c) If f and g are bijections from A to A, then so are $g \circ f$ and $f \circ g$.

d) If f, g, h are bijective functions from A to A and $f \circ g = f \circ h$, then $g=h$.

Apply $f^{-1}$ at left to both sides to obtain $g=h$.

Coming to the problem at hand, since A has n elements, we see that the there are only finitely many (in fact, n!) bijective functions from A to A as each bijective function f gives a permutation of $\{ 1,2,3,\ldots, n\}$ by taking $\{ f(1),f(2), \ldots, f(n)\}$. Since f is a bijective function from A to A, so is each of the function in the sequence:

$f^{2}, f^{3}, \ldots, f^{n}, \ldots$

All these cannot be distinct, since there are only finitely many bijective functions from A to A. Hence, for some two distinct positive integers m and n, $m > n$, say, we must have $f^{m}=f^{n}$

If $n=1$, we take $M=m$, to obtain the result. If $n>1$, multiply both sides by $(f^{-1})^{n-1}$ to get $f^{m-n+1}=f$. We take $M=m-n+1$ to get the relation $f^{M}=f$ with $M>1$. Note that this means $f^{M}(i)=f(i)$ for all $i \in A$. QED.

Problem 6:

Show that there exists a convex hexagon in the plane such that :

a) all its interior angles are equal

b) its sides are 1,2,3,4,5,6 in some order.

Solution 6:

Let ABCDEF be an equiangular hexagon with side lengths 1,2,3,4,5,6 in some order. We may assume without loss of generality that $AB=1$. Let $BC=a, CD=b, DE=c, EF=d, FA=e$.

Since the sum of all angles of a hexagon is equal to $(6-2) \times 180=720 \deg$, it follows that each interior angle must be equal to $720/6=120\deg$. Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. We use the vector method: if the vector is denoted by $(x,y)$ we then have:

$\overline{AB}=(1,0)$

$\overline{BC}=(a\cos 60\deg, a\sin 60\deg)$

$\overline{CD}=(b\cos{120\deg},b\sin{120\deg})$

$\overline{DE}=(c\cos{180\deg},c\sin{180\deg})=(-c,0)$

$\overline{EF}=(d\cos{240\deg},d\sin{240\deg})$

$\overline{FA}=(e\cos{300\deg},e\sin{300\deg})$

This is because these vectors are inclined to the positive x axis at angles 0, 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300 degrees respectively.

Since the sum of all these six vectors is $\overline{0}$, it implies that $1+\frac{a}{2}-\frac{b}{2}-c-\frac{d}{2}+\frac{e}{2}=0$

and $(a+b-d-e)\frac{\sqrt{3}}{2}=0$

That is, $a-b-2c-d+e+2=0$….call this I

and $a+b-d-e=0$….call this II

Since $(a,b,c,d,e)=(2,3,4,5,6)$, in view of (II), we have

$(a,b)=(2,5), (a,e)=(3,4), c=6$….(i)

$(a,b)=(5,6), (a,e)=(4,5), c=2$…(ii)

$(a,b)=(2,6), (a,e)=(5,5), c=4$…(iii)

The possibility that $(a,b)=(3,4), (a,e)=(2,5)$ in (i), for instance, need not be considered separately, because we can reflect the figure about $x=\frac{1}{2}$ and interchange these two sets.

Case (i):

Here $(a-b)-(d-e)=2c-2=10$. Since $a-b=\pm 3, d-e=\pm 1$, this is not possible.

Case (ii):

Here $(a-b)-(d-e)=2c-2=2$. This is satisfied by $(a,b,d,e)=(6,3,5,4)$

Case (iii):

Here $(a-b)-(d-e)=2c-2=6$

Case (iv):

This is satisfied by $(a,b,d,e)=(6,2,3,5)$.

Hence, we have (essentially) two different solutions: $(1,6,3,2,5,4)$ and $(1,6,2,4,3,5)$. It may be verified that I and II are both satisfied by these sets of values.

Aliter: Embed the hexagon in an appropriate equilateral triangle, whose sides consist of some sides of the hexagon.

Solutions to the remaining problems from that blog will have to be tried by the student.

Cheers,

Nalin Pithwa.

Miscellaneous questions: part II: tutorial practice for preRMO and RMO

Problem 1:

Let $a_{1}, a_{2}, \ldots, a_{10}$ be ten real numbers such that each is greater than 1 and less than 55. Prove that there are three among the given numbers which form the lengths of the sides of a triangle.

Problem 2:

In a collection of 1234 persons, any two persons are mutual friends or enemies. Each person has at most 3 enemies. Prove that it is possible to divide this collection into two parts such that each person has at most 1 enemy in his subcollection.

Problem 3:

A barrel contains 2n balls numbered 1 to 2n. Choose three balls at random, one after the other, and with the balls replaced after each draw. What is the probability that the three element sequence obtained has the properties that the smallest element is odd and that only the smallest element, if any, is repeated?

That’s all, folks !!

You will need to churn a lot…!! In other words, learn to brood now…learn to think for a long time on a single hard problem …

Regards,
Nalin Pithwa

Miscellaneous questions: Part I: tutorial practice for preRMO and RMO

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 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 sixty four entries are in fact equal.

Problem 2:

Let T be the set of all triples (a,b,c) of integers such that $1 \leq a < b < c \leq 6$. For each triple (a,b,c) in T, take the product abc. Add all these products corresponding to all triples in I. Prove that the sum is divisible by 7.

Problem 3:

In a class of 25 students, there are 17 cyclists, 13 swimmers, and 8 weight lifters and no one in 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.

Problem 4:

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 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.

Problem 5:

Let f be a bijective (one-one and onto) function from the set $A=\{ 1,2,3,\ldots,n\}$ to itself. Show that there is a positive integer $M>1$ such that $f^{M}(i)=f(i)$ for each $i \in A$. Note that $f^{M}$ denotes the composite function $f \circ f \circ f \ldots \circ f$ repeated M times.

Problem 6:

Show that there exists a convex hexagon in the plane such that:
a) all its interior angles are equal
b) its sides are 1,2,3,4,5,6 in some order.

Problem 7:

There are ten objects with total weights 20, each of the weights being a positive integer. Given that none of the weights exceed 10, prove that the ten objects can be divided into two groups that balance each other when placed on the pans of a balance.

Problem 8:

In each of the eight corners of a cube, write +1 or -1 arbitrarily. Then, on each of the six faces of the cube write the product of the numbers written at the four corners of that face. Add all the fourteen numbers so writtein down. Is it possible to arrange the numbers +1 and -1 at the corners initially so that this final sum is zero?

Problem 9:

Given the seven element set $A = \{ a,b,c,d,e,f,g\}$ find a collection T of 3-element subsets of A such that each pair of elements from A occurs exactly in one of the subsets of T.

Try these !!

Regards,
Nalin Pithwa

A fifth degree equation in two variables: a clever solution

Question:

Verify the identity: $(2xy+(x^{2}-2y^{2}))^{5}+(2xy-(x^{2}-2y^{2}))^{5}=(2xy+(x^{2}+2y^{2})i)^{5}+(2xy-(x^{2}+2y^{2})i)^{5}$

let us observe first that each of the fifth degree expression is just a quadratic in two variables x and y. Let us say the above identity to be verified is:

$P_{1}+P_{2}=P_{3}+P_{4}$

Method I:

Use binomial expansion. It is a very longish tedious method.

Method II:

Factorize each of the quadratic expressions $P_{1}, P_{2}, P_{3}, P_{4}$ using quadratic formula method (what is known in India as Sridhar Acharya’s method):

Now fill in the above details.

You will conclude very happily that :

The above identity is transformed to :

$P_{1}=(x+y+\sqrt{3}y)^{5}(x+y-\sqrt{3}y)^{5}$

$P_{2}=(-1)^{5}(x-y-\sqrt{3}y)^{5}(x-y+\sqrt{3}y)^{5}$

$P_{3}=(i^{2}(x-y-\sqrt{3}y)(x-y+\sqrt{3}y))^{5}$

$P_{4}=((-i^{2})(x+y+\sqrt{3}y)(x-y-\sqrt{3}y))^{5}$

You will find that $P_{1}=P_{4}$ and $P_{2}=P_{4}$

Hence, it is verified that the given identity $P_{1}+P_{2}=P_{3}+P_{4}$. QED.

Regards,
Nalin Pithwa.

A quadratic equation question for pRMO or preRMO

Question:

Find the necessary and sufficient condition that the quadratic equation $ax^{2}+bx+c=0$ where $a \neq 0$ has one root which is the square of the other.

Solution:

Let the two roots of the given quadratic equation $ax^{2}+bx+c=0$, with $a \neq 0$ be $\alpha$ and $\beta$ such that $\beta = \alpha^{2}$.

Then, we know $\alpha+\beta=-\frac{b}{a}$ and $\alpha\beta=\frac{c}{a}$ so that $\alpha+\alpha^{2}=-\frac{b}{a}$ and $\alpha^{3}=\frac{c}{a}$. From the latter relation, we get that $\alpha = (\frac{c}{a})^{\frac{1}{3}}$. Substituting this in the first relation of sum of roots, we get the following necessary and sufficient condition:

$(\frac{c}{a})^{\frac{1}{3}} + (\frac{c}{a})^{\frac{2}{3}} = -\frac{b}{a}$.

The above is the desired solution.

🙂 🙂 🙂

Nalin Pithwa