# An optimization problem in geometry — RMO and INMO

An exerciser wants to twirl a 1 meter long baton in a horizontal plane through 360 degrees as he moves around in a room without hitting the walls. He does not mind any shape but wants to minimize the area of the room. Note that a circular room of 1 meter diameter (with an area of pi/4 square meters) is far from the minimum area possible.

Pick up this challenge for yourself! Victory is ours when we have strength and courage to run our own race!

Nalin Pithwa.

# Geometry and Trigonometry Toolkit — RMO

Geometry:

1. The area of two triangles having equal bases (heights) are in the ratio of their heights (bases).
2. If ABC and DEF are two triangles, then the following statements are equivalent: (a) $\angle {A}=\angle{D}$, $\angle{B}=\angle{E}$, $\angle{C}=\angle{F}$ (b) $\frac{BC}{EF} = \frac{CA}{FD} = \frac{AB}{DE}$ (c) $\frac{AB}{AC} = \frac {DE}{DF}$ and $\angle {A} = \angle {D}$ Two triangles satisfying any one of these conditions are said to be similar to each other.
3. Appolonius Theorem: If D is the mid-point of the side BC in a triangle ABC then $AB^{2}+AC^{2}=2(AD^{2}+BD^{2})$.
4. Ceva’s Theorem: If ABC is a triangle, P is a point in its plane and AP, BP, CP meet the sides BC, CA, AB in D, E, F respectively, then $\frac{BD}{DC}. \frac{CE}{EA}. \frac{AF}{FB} = +1$. Conversely, if D, E, F are points on the (possibly extended) sides BC, CA, AB respectively and the above relation holds good, then AD, BE, CF concur at a point. Lines such as AD, BE, CF are called Cevians.
5. Menelaus’s Theorem: If ABC is a triangle and a line meets the sides BC, CA, AB in D, E, F respectively, then $\frac{BD}{DC}. \frac{CE}{EA}. \frac{AF}{FB} = -1$, taking directions of the line segments into considerations, that is, for example, $CD = -DC$. Conversely, if on the sides BC, CA, AB (possibly extended) of a triangle ABC, points D, E, F are taken respectively such that the above relation holds good, then D, E, F are collinear.
6. If two chords AB, CD of a circle intersect at a point O (which may lie inside or outside the circle), then $AO.OB = CO.OD$. Conversely, if AB and CD are two line segments intersecting at O, such that $AO.OB=CO.OD$, then the four points A, B, C, D are concyclic.
7. (This may be considered as a limiting case of 6, in which A and B coincide and the chord AB becomes the tangent at A). If OA is a tangent to a circle at A from a point O outside the circle and OCD is any secant of the circle (that is, a straight line passing through O and intersecting the circle at C and D), then $OA^{2}=OC.OD$. Conversely, if OA and OCD are two distinct line segments such that $OA^{2}=OC.OD$, then OA is a tangent at A to the circumcircle of triangle ABC.
8. Ptolemy’s Theorem: If ABCD is a cyclic quadrilateral, then $AB.CD + AD.BC = AC.BD$. Conversely, if in a quadrilateral ABCD this relation is true, then the quadrilateral is cyclic.
9. If AB is a line segment in a plane, then set of points P in the plane such that $\frac{AP}{PB}$ is a fixed ratio $\lambda$ (not equal to 0 or 1) constitute a circle, called the Appolonius circle. If C and D are two points on AB dividing the line segment AB in the ratio $\lambda : 1$ internally and externally, then C and D themselves are two points on the circle such that CD is a diameter. Further, for any points P on the circle, PC and PD are the internal and external bisectors of $\angle{APB}$.
10. Two plane figures $\alpha$ and $\beta$ such as triangles, circles, arcs of a circle are said to be homothetic relative to a point O (in the plane) if for every point A on $\alpha$, OA meets $\beta$ in a point B such that $\frac{OA}{OB}$ is a fixed ratio $\lambda$ (not equal to zero). The point O is called the centre of similitude or hometheity. Also, any two corresponding points X and Y of the figures $\alpha$ and $\beta$ (example, the circumcentres of two homothetic triangles) are such that O, X, Y are collinear and $\frac{OX}{OY}=\lambda$.

Trigonometry:

1. Compound and Multiple Angles:
• $\sin{(A \pm B)} = \sin{A}\cos{B} \pm \cos{A}\sin{B}$; $\cos{(A \pm B)} = \cos{A}\cos{B} \mp \sin{A}\sin{B}$ ; $\tan{(A \pm B)} = \frac{(\tan{A} \pm \tan{B})}{1 \mp \tan{A}\tan{B}}$.
• $\sin{2A} = 2\sin{A}\cos{A} = \frac{2\tan{A}}{(1+\tan^{2}{A})}$; $\cos{2A}=\cos^{2}{A}-\sin^{2}{A}=2\cos^{2}{A}-1=1-2\sin^{2}{A}=\frac{1-\tan^{2}{A}}{1+\tan^{2}{A}}$; $\tan{2A}=\frac{2\tan{A}}{1-\tan^{2}{A}}$
• $\sin{3A}=3\sin{A}-4\sin^{3}{A}$; $\cos{3A}=4\cos^{3}{A}-3\cos{A}$; $\tan{3A}=\frac{(3\tan{A}-\tan^{3}{A})}{1-3\tan^{2}{A}}$;
2. Conversion Formulae:
1. $\sin{C} + \sin {D} = 2\sin{(\frac{C+D}{2})}\cos{(\frac{C-D}{2})}$
2. $\sin{C} - \sin {D} = 2\cos{(\frac{C+D}{2})}\sin{(\frac{C-D}{2})}$
3. $\cos{C} + \cos{D} = 2\cos{(\frac{C+D}{2})}\cos{(\frac{C-D}{2})}$
4. $\cos{C} - \cos{D} = 2\sin{(\frac{C+D}{2})}\sin{\frac{D-C}{2}}$
5. $2\sin{A}\cos{B}= \sin{(A+B)} + \sin{(A-B)}$
6. $2\cos{A}\sin{B} = \sin{(A+B)} - \sin{(A-B)}$
7. $2\cos{A}\cos{B} = \cos{(A+B)} + \cos{(A-B)}$
8. $2\sin{A}\sin{B} = \cos{(A-B)} - \cos {(A+B)}$
3. Properties of Triangles:
1. Sine Rule: $\frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}} = 2R$
2. Cosine Rule: $a^{2}=b^{2}+c^{2}-2bc\cos{A}$
3. Half-Angle Rule: $\sin{\frac{A}{2}} = \sqrt{\frac{(s-b)(s-c)}{bc}}$; $\cos{\frac{A}{2}} = \sqrt{\frac{s(s-a)}{bc}}$; $\tan{\frac{A}{2}} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$
4. Circumradius: $R = \frac{abc}{4(area)}$
5. In-radius: $r=4R\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}= (s-a)\tan{\frac{A}{2}}$
6. Area: $\Delta = rs= \frac{1}{2}bc\sin{A}=2R^{2}\sin{A}\sin{B}\sin{C}=\frac{abc}{4R}=\sqrt{s(s-a)(s-b)(s-c)}$
7. Medians: $m_{a}=\frac{1}{2}\sqrt{2b^{2}+2c^{2}-a^{2}}$ and similar expressions for other medians.
4. Miscellaneous:
1. $a=b\cos{C}+c\cos{B}$, $b=a\cos{C}+c\cos{A}$, $c=a\cos{B} + b\cos{A}$
2. If O is the circumcentre and X is the mid-point of BC, then $\angle{BOX} = \angle{COX} = A$ and $OX = R\cos{A}$.
3. If AD is the altitude with D on BC and H the orthocentre, then $AH = 2R \cos{A}$, $HD = 2R\cos{B}\cos{C}$.
4. If G is the centroid and N the nine-point centre, then O, G, N, H are collinear and $OG:GH = 1:2$, $ON = NH$.
5. If I is the in-centre, then $\angle{BIC} = 90 \hspace{0.1in}degrees + \frac{A}{2}$
6. The centroid divides the medians in the ratio $2:1$
7. $OI^{2}=R^{2}-2Rr=R^{2}(1-8\sin{A/2}\sin{B/2}\sin{C/2})$; $OH^{2}=R^{2}(1-8\cos{A}\cos{B}\cos{C})=9R^{2}-a^{2}-b^{2}-c^{2}$; $HI^{2}=2r^{2}-4R^{2}\cos{A}\cos{B}\cos{C}$
8. If $a+b+c=\pi$, then $\sin{A}+\sin{B}+\sin{C}=4\cos{A/2}\cos{B/2}\cos{C/2}$; $\cos{A} + \cos{B} + \cos{C} = 1 + 4\sin{A/2}\sin{B/2}\sin{C/2}$; $\tan{A} + \tan{B} + \tan{C} = \tan{A}\tan{B}\tan{C}$; $\sin{2A}+\sin{2B} + \sin{2C} = 4\sin{A}\sin{B}\sin{C}$; $\cos^{2}{A} + \cos^{2}{B} + \cos^{2}{C} = 1 + 2\cos{A}\cos{B}\cos{C}$
9. Area of a quadrilateral ABCD with $AB=a$; $BC=b$; $CD=c$; $DA=d$, $A+C=2\alpha$ is given by $\Delta = \sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^{2}\alpha}$ If it is cyclic, then $\Delta = \sqrt{(s-a)(s-b)(s-c)(s-d)}$. Its diagonals are given by $AC = \sqrt{\frac{(ac+bd)(ad+bc)}{(ab+cd)}}$; $BD = \sqrt{\frac{(ac+bd)(ab+cd)}{(ad+bc)}}$.

Some important comments: To the student readers: from Ref: Problem Primer for the Olympiad by C. R. Pranesachar, B. J. Venkatachala, C. S. Yogananda: The best way to use this book is, of course, to look up the problems and solve them!! If you cannot get started then look up the section “Tool Kit”, which is a collection of theorems and results which are generally not available in school text books, but which are extremely useful in solving problems. As in any other trade, you will have to familiarize yourself with the tools and understand them to be able to use them effectively. We strongly recommend that you try to devise your own proofs for these results of Tool Kit or refer to other classic references. …

My remark: This is the way to learn math from scratch or math from first principles.

The book is available in Amazon India or Flipkart:

Amazon India:

Flipkart:

More geometry later!

Nalin Pithwa

# RMO (Regional Mathematics Olympiad) — Plane Geometry Basics

Let us get started on plane geometry, some how or the other, now. Rather than starting with a lecture, let me start in a un-conventional format. A little challenging problem for you. Try to prove the following (Morley’s theorem):

In any arbitrary triangle, draw the trisectors of all the angles. Prove that the triangle formed by the points of intersection of adjacent trisectors (of different angles), taken in pair, is always equilateral.

Important Note: Please do not take look at any solution from the internet or any other source. You can use the basic theorems only presented in the Geometry toolkit, like as in, Problem Primer for the Olympiad, B J Venkatachala and/or in the lecture.

Have fun!

Nalin Pithwa.

# Another Romanian Mathematical Olympiad problem

Ref: Romanian Mathematical Olympiad — Final Round, 1994

Ref: Titu Andreescu

Problem:

Let M, N, P, Q, R, S be the midpoints of the sides AB, BC, CD, DE, EF, FA of a hexagon. Prove that

$RN^{2}=MQ^{2}+PS^{2}$ if and only if MQ is perpendicular to PS.

Proof:

Let a, b, c, d, e, f be the coordinates of the vertices of the hexagon. The points M, N, P, Q, R, and S have the coordinates

$m=\frac{a+b}{2}$, $n=\frac{b+c}{2}$, $=\frac{c+d}{2}$,

$q=\frac{d+e}{2}$, $r=\frac{e+f}{2}$, $s=\frac{f+a}{2}$, respectively.

Using the properties of the real product of complex numbers, (please fill in the gaps here), we have

$RN^{2}=MQ^{2}+PS^{2}$

if and only if

$(e+f-b-c).(e+f-b-c) = (d+e-a-b).(d+e-a-b)+(f+a-c-d).(f+a-c-d)$

That is,

$(d+e-a-b).(f+a-c-d)=0$

hence, MQ is perpendicular to PS, as claimed. QED.

More later,

Nalin Pithwa

# A geometry problem for the Regional Mathematical Olympiad

Problem:

Prove that in any acute angled triangle of sides a, b, c, semi perimeter p, in-radius r, and circumradius R, the following inequalities hold:

$\frac{2}{5} \leq \frac{Rp}{2aR+bc} < \frac{1}{2}$

Proof:

Let D be the foot of the altitude from A, and D will be on the side BC since the triangle has only acute angles. Now, by summing up $AD+BD>AB$ and $CD+AD>AC$, we get

$2h_{a}+a>b+c \Longrightarrow h_{a}>p-a \Longrightarrow \frac{2S}{a}>p-a \Longrightarrow 2S > a(p-a) \Longrightarrow \frac{abc}{2R} > a(p-a) \Longrightarrow bc+2aR > 2pR \Longrightarrow \frac{pR}{bc+2aR}<\frac{1}{2}$

For the other part, we have the following equivalences:

$\frac{2}{5} \leq \frac{Rp}{2aR+bc} \Longleftrightarrow 2aR+bc \leq \frac{5pR}{2} \Longleftrightarrow 4aR + \frac{8SR}{a} \leq 5pR \Longleftrightarrow 4a^{2}+4ah_{a} \leq 5ap \Longleftrightarrow (a+h_{a}) \leq \frac{5p}{4}$

—– call the above as relation I

But, $b=\sqrt{h_{a}^{2}+BD^{2}}$ and $c=\sqrt{h_{a}^{2}+CD^{2}}$ and by Minkowski’s inequality, we have $b+c \geq \sqrt{4h_{a}^{2}+a^{2}}$. Then, we will have

$\frac{5p}{4} = \frac{5(a+b+c)}{8} \geq \frac{5(a+\sqrt{4h_{a}^{2}+a^{2}})}{8} \geq a+h_{a} \Longleftrightarrow 5\sqrt{4h_{a}^{2}+a^{2}} \geq 3a +8h_{a} \Longleftrightarrow 100h_{a}^{2}+25a^{2} \geq 64h_{a}^{2}+9a^{2}+48ah_{a} \Longleftrightarrow 36h_{a}^{2}+16a^{2} \geq 48ah_{a}$

which holds for all positive a and $h_{a}$ because of the AM-GM inequality. This tells us that (1) is true, and thus so is our conclusion.

Ref: Problems for the Mathematical Olympiads (from the First Team Selection Test to the IMO) by Andrei Negut

More geometry later !

Nalin Pithwa

# Fermat’s numbers and infinity of primes

There are two extremely beautiful, fundamental theorems in mathematics — (1) the number of primes is infinite (2) the square root of 2 is irrational.

The oldest proof of the first theorem is due to Euclid. There is another proof of the first theorem due to Christian Goldbach (from a letter to Leonhard Euler, 1730). It is as follows: (Reference: Proofs from THE BOOK by Martin Aigner and Gunter M. Ziegler, fourth edition)

Let us look at Fermat Numbers $F_{n}=2^{2^{n}}+1$ for $n=0, 1, 2, \ldots$. We will show that any two Fermat numbers are relatively prime; hence, there must be infinitely many primes. To this end, we verify the recursion

$\Pi_{k=0}^{n-1}F_{k}=F_{n}-2$ for $n \geq 1$

from which our assertion follows immediately. Indeed, if m is a divisor of $F_{k}$ and $F_{n}$ ($k < n$), then m divides 2, and, hence, $m =1 \hspace{0.1in} or \hspace{0.1in} 2$. But, m=2 is impossible since all Fermat numbers are odd.

To prove the recursion, we use induction on n. For $n=1$, we have $F_{0}=3$ and $F_{1}-2=3$. With induction, we now conclude:

$\Pi_{k=0}^{n}F_{k}=(\Pi_{k=0}^{n-1}F_{k})F_{n}=(F_{n}-2)F_{n}=(2^{2^{n}}-1)(2^{2^{n}}+1)$, which in turn, equals

$2^{2^{n+1}}-1=F_{n+1}-2$. QED.

More later,

Nalin Pithwa

# Geometric mean and a basic theorem

MEANS.

The average of two lengths (or numbers) is called their arithmetic mean: $\frac{1}{2}(a+b)$ is the arithmetic mean of a and b. The geometric mean is the square root of their product: $\sqrt{ab}$. The arithmetic mean of 8 and 2 is 5; their geometric mean is 4.

It is easy to find the geometric mean of two positive numbers algebraically, by solving for x the equation $a/x=x/b$, for  this says $x^{2}=ab$. For this reason, the geometric mean is also called the mean proportional between a and b.

Is there any way to find these quantities with ruler and compass? The arithmetic mean is easy enough: simply lay off the  two lengths end to end on the same straight line and bisect the total segment. A method for the geometric mean is suggested by the equation: $\frac{AB}{x}=\frac{x}{BC}$.

Using $AB+BC$ as diameter, draw a circle and the chord perpendicular to  that diameter at B. (Fig 1). This chord is bisected by the diameter, and we recognize a special case of Theorem 2,

namely, $x.x=AB.BC$, which says that x is the mean proportional between AB and BC.

In practice, one uses only half the figure. (Fig 2).

Inasmuch as MO, the perpendicular from O, the mid-point of AC, is the longest perpendicular to the semicircle from its diameter, and is also a radius equal to $\frac{AC}{2}=\frac{AB+BC}{2}$, we have:

THEOREM 3.

The geometric mean of two unequal positive numbers is always less than their arithmetic mean.

Why do we need the word “unequal”?

We now restate the part of Theorem 2,

THEOREM 4.

If, from an external point, a tangent and a secant are drawn, the tangent is the mean proportional between the whole secant and its external segment.

Of course, the two segments PA and PB become the same length when the chord is moved into the limiting position of tangency, bringing A and B together. If you are bothered by this, note that there is a proof possible, which is quite independent of any limiting process.

if we relabel the diagram as in Fig 3 here, and then call OT by the single letter t, we have

$t^{2}=OC.OD$

if and only if t is a tangent. These labels are chosen to fit the lettering of the next blog on this topic.

I trust you will agree that we have done nothing difficult so far. Yet you may soon be surprised at the structure we are about to build on this small but sturdy foundation. We are now ready to look at some of the geometry they didn’t teach you.

More later,

Nalin Pithwa