# Pure plane geometry problems for RMO training or practice

Similar Figures:

Definition:

Polygons which are equiangular and have their corresponding sides proportional are called similar.

If, in addition, their corresponding sides are parallel, they are said to be similarly situated or homothetic.

Theorem 1:

If O is any fixed point and ABCD…X any polygon, and if points $A^{'}, B^{'}, C^{'}, \ldots X^{'}$ are taken on OA, OB, OC, …OX (or those lines produced either way) such that $\frac{OA^{'}}{OA} = \frac{OB^{'}}{OB} = \ldots = \frac{OX^{'}}{OX}$, then the polygons $ABCD...X$ and $A^{'}B^{'}C^{'} \ldots X^{'}$ are homothetic.

Before we state the next theorem, some background is necessary.

If O is a fixed point and P is a variable point on a fixed curve S, and if $P^{'}$ is a point on OP such that $\frac{OP^{'}}{OP} = k$, a constant, then the locus of $P^{'}$ is a curve $S^{'}$, which is said to be homothetic to S; and $P, P^{'}$ are corresponding points.

O is called the centre of similitude of the two figures.

If $P$ and $P^{'}$ lie on the same side of O, the figures are said to be directly homothetic w.r.t.O, and O is called the external centre of similitude.

If $P$ and $P^{'}$ lie on the opposite sides of O, the figures are said to be inversely homothetic w.r.t. O, and O is called the internal centre of similitude.

If we join A and B in the first case, we say that the parallel lines $AB, A^{'}B^{'}$ are drawn in the same sense, and in the second case, in opposite senses.

Theorem 2:

Let A, B be the centres of any two circles of radii a, b; AB is divided externally at O and internally at $O_{1}$ in the ratio of the radii, that is, $\frac{AO}{BO} = \frac{AO_{1}}{O_{1}B} = \frac{a}{b}$ then the circles are directly homothetic w.r.t. O and inversely homothetic w.r.t. $O_{2}$, and corresponding points lie on the extremities of parallel radii.

Now, prove the above two hard core basic geometry facts. ! 🙂 🙂 🙂

Cheers,

Nalin Pithwa.

# 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