# Pick’s theorem: a geometry problem for RMO practice

Pick’s theorem:

Consider a square lattice of unit side. A simple polygon (with non-intersecting sides) of any shape is drawn with its vertices at the lattice points. The area of the polygon can be simply obtained as $B/2+I-1$ square units, where B is the number of lattice points on the boundary; I = number of lattice points in the interior region of the polygon. Prove this theorem.

Proof:

Refer Wikipedia 🙂 🙂 🙂

https://en.wikipedia.org/wiki/Pick%27s_theorem

Cheers,

Nalin Pithwa.

# A “primer” on geometric inequalities for pre-RMO and RMO

The comparison of lengths is more basic than comparison of other geometric quantities (such as angles, areas and volumes). A geometric inequality that involves only the lengths is called a distance inequality.

Some simple axioms and theorems on inequalities in Euclidean geometry are usually the starting point to solve the problem of distance inequality, in which most frequently used tools are:

Proposition I:

The shortest line connecting point A with point B is the segment AB.

The direct corollary of proposition 1 is as follows:

Proposition 2:

(Triangle inequality)

For arbitrary three points A, B and C (lying in the same plane), we have $AB \leq AC +CB$, the equality holds iff the three points are collinear.

Remark:

In most literature, any symbol of a geometric object also denotes its quantity according to the context.

Proposition 3:

In a triangle, the longer side has the larger opposite angle. And, conversely, the longer angle has the longer opposite side.

Proposition 4:

The median of a triangle on a side is shorter than the half-sum of the other two sides.

Proposition 5:

If a convex polygon is within another one, then the outside convex polygon’s perimeter is larger.

Proposition 6:

Any segment in a convex polygon is either less than the longest side or the longest diagonal of the convex polygon.

Here, is a classic example:

Example 1:

Let $a, b, c$ be the sides of a $\triangle ABC$. Prove that $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}<2$.

Solution 1:

By the triangle inequality, $a yields $\frac{a}{b+c} = \frac{2a}{2(b+c)} < \frac{2a}{a+b+c}$

Similarly, $\frac{b}{c+a} < \frac{2b}{a+b+c}$ and $\frac{c}{b+a} < \frac{2c}{a+b+c}$

Adding up the above three inequalities, we get the required inequality.

Example 2:

Let AB be the longest side of $\triangle ABC$, and P a point in the triangle, prove that $PA+PB>PC$.

Solution/Proof 2:

Let D be the intersection point of CP and AB. (Note P is in the interior of the triangle ABC). Then, $\angle ADC$ or $\angle BDC$ is not acute. Without loss of generality, we assume that $\angle ADC$ is not acute. Applying proposition 3 to $\triangle ADC$, we obtain $AC \geq CD$. Therefore, $AB \geq AC \geq CD > PC$….call this as  relationship “a”.

Furthermore, applying triangle inequality to $\triangle PAB$, we have $PA+PB>AB$…call this as relationship “b”.

Combining “a” and “b”, we obtain the required inequality immediately. QED.

Remarks: (1) If AB is not the longest, then the conclusion may not be true. (2) If point P on the plane of regular triangle ABC, P is not on the circumcircle of the triangle, then the sum of any two of PA, PB, and PC is longer than the remaining one. That is, PA, PB and PC consist of a triangle’s three sides.

Quiz:

Prove that a closed polygonal line with perimeter 1 can be put inside a circle with radius 0.25.

Reference:

Geometric Inequalities, Vol 12, Gangsong Leng, translated by Yongming Liu.

Cheers,

Nalin Pithwa.

# RMO Geometry Basics: Solutions to Bertschneider/Brahmagupta’s formulae

Well, the solutions already exist ! (pun intended! 🙂 🙂 :-))

You may note that putting one of  the sides of a quadrilateral to zero (thereby reducing it to a triangle), one recovers Heron’s formula. Consider the quadrilateral as a combination of two triangles by drawing one of the diagonals. The length of the diagonal can be expressed in terms of the lengths of the sides and (cosine of) two diagonally opposite angles. Then, use the Heron’s formula for each of the triangles. Through algebraic manipulation, one can get the required result. If necessary, the reader is advised to consult again Wikipedia Mathematics on the internet!

🙂 🙂 🙂

Nalin Pithwa.

# RMO Geometry : Basics : Bertschneider (Coolidge)/Brahmagupta’s Formula

Heron’s formula for the area of a triangle is well-known. A similar formula for the area of a quadrilateral in terms of the lengths of its sides is given below:

Note that the lengths of the four sides do not specify the quadrilateral uniquely.The area

$\Delta=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd.cos^{2}(\phi/2)}$

where a, b, c, and d are the lengths of the four sides; s is the semi-perimeter and $\phi$ is the sum of the diagonally opposite angles of the quadrilateral. This is known as Bertschneider(Coolidge) formula. For a cyclic quadrilateral, $\phi$ is 180 degrees and the area is maximum for the set of given sides and the area is given by (Brahmagupta’s formula):

$\Delta = \sqrt{(s-a)(s-b)(s-c)(s-d)}$.

Prove both the formulae given above!

-Nalin Pithwa.

PS: I will put the solutions on this blog after some day(s). First, you need to try.

# Cyclic quadrilaterals — Plane geometry for RMO

A convex quadrilateral is called cyclic if its vertices lie on a circle. It is not difficult to see that a necessary and sufficient condition for this is that the sum of the opposite angles of the quadrilateral be equal to 180 degrees.

As a special case, if two opposite angles of the quadrilateral are right angles, then the quadrilateral is cyclic and one of its diagonals is a diameter of the circumscribed circle.

Another necessary and sufficient condition is that the angle between one side and a diagonal be equal to the angle between the opposite side and the other diagonal.

Problem:

Let ABCD be a cyclic quadrilateral. Recall that the incenter of a triangle is the intersection of the angles’ bisectors. Prove that the incenters of triangles ABC, BCD, CDA and DAB are the vertices of a rectangle.

Comment: It is easy. Give it a shot!

-Nalin Pithwa.

# Optimization problems in Geometry — RMO training

Problem 1:

Within a given triangle ABC having all angles less than 120 degrees, determine the point P, so that $PA+PB+PC$ is minimum.

Problem 2:

Within a given convex quadrilateral ABCD, determine the point P, so that $PA+PB+PC+PD$ is minimum.

Problem 3:

In an acute-angled triangle ABC, determine the points D on AB, E on BC, and F on AC so that the perimeter of the triangle DEF is minimum.

Problem 4:

Three cities are located on the vertices of an equilateral triangle of sides 100 km. What must be the minimum total length of the roads connecting these cities so that one can travel from any city to another?

Problem 5:

Four cities are located on the vertices of a square of sides 100 km. What must be the minimum length of the roads connecting these cities so that one can travel from any city to another?

Problem 6:

Consider a park of quadrilateral shape ABCD. A house is located at P on the edge AB. Three more houses are to be built at Q on the edge AD, at R on the edge CD and at S on the edge CB. Locate the points Q, R and S so that the total length of  the road PQRS directly connecting these four houses, constructed within the park, is minimized.

Have some fun with geometry now !

Nalin Pithwa

# More problems in pure plane geometry for RMO

Problem 1:

A triangle is divided into two parts by drawing a line through the centroid. Prove that the area of the smaller part is at least 80 % of the bigger part. In fact, this statement is true for all convex figures and is known as Winternitz theorem.

Problem 2:

In a rectangle ABCD, the side $AB>BC$. Locate geometrically (use of only a compass and an unmarked straightedge is allowed) the points X and Y on CD, so that $AX=XY=YB$.

Problem 3:

In a triangle ABC, $AB/2. Locate geometrically the points D on AB and E on AC, so that $BD=DE=EC$.

Problem 4:

P is a point inside a square ABCD such that $\angle{PCD}=\angle{PDC}=15 degrees$. Prove that the triangle PAB is equilateral.

Problem 5:

Four circles are drawn, all of same radius r and passing through a point O. Let the quadrilateral ABCD consisting of direct tangents to this set of circles be the circumscribing quadrilateral. Prove that ABCD is a cyclic quadrilateral.

Problem 6:

Viviani’s Theorem: Prove that the sum of the distances of a point inside an equilateral triangle from the three sides is independent of the point.

Problem 7:

Geometrically construct a lune (a concave area bounded by two circular arcs) of unit area.

Have fun!

Nalin Pithwa.

# Solidify your plane geometry for RMO: Archimedes Problem

Solve the following historically famous problem (Archimedes):

On a circular arc AB, the point M is midway, that is, arc AM is equal to arc MB. The point C is on the arc MB. The point D is the foot of the perpendicular from M onto the line AC. Prove that D is the mid-point of the straight line path $AC + CB$.

Nalin Pithwa.

# 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: