# RMO and Pre RMO Geometry Tutorial Worksheet 1: Based on Geometric Refresher

1) Show that quadrilateral ABCD can be inscribed in a circle iff $\angle B$ and $\angle D$ are supplementary.

2) Prove that a parallelogram having perpendicular diagonals is a rhombus.

3) Prove that a parallelogram with equal diagonals is a rectangle.

4) Show that the diagonals of an isosceles trapezoid are equal.

5) A straight line cuts two concentric circles in points A, B, C and D in that order. AE and BF are parallel chords, one in each circle. If CG is perpendicular to BF and DH is perpendicular to AE, prove that $GF = HE$.

6) Construct triangle ABC, given angle A, side AC and the radius r of the inscribed circle. Justify your construction.

7) Let a triangle ABC be right angled at C. The internal bisectors of angle A and angle B meet BC and CA at P and Q respectively. M and N are the feet of the perpendiculars from P and Q to AB. Find angle MCN.

8) Three circles $C_{1}, C_{2}, C_{3}$ with radii $r_{1}, r_{2}, r_{3}$, with $r_{1}. They are placed such that $C_{2}$ lies to the right of $C_{1}$ and touches it externally; $C_{3}$ lies to the right of $C_{2}$ and touches it externally. Further, there exist two straight lines each of which is a direct common tangent simultaneously to all the three circles. Find $r_{2}$ in terms of $r_{1}$ and $r_{3}$.

Cheers,

Nalin Pithwa

# Basic Geometry: Refresher for pre-RMO, RMO and IITJEE foundation maths

1) Assume that in a $\bigtriangleup$ ABC and $\bigtriangleup RST$, we know that $AB=RS$, $AC=RT$, $BC=ST$. Prove that $\bigtriangleup ABC \cong \bigtriangleup RST$ without using the SSS congruence criterion.

2) Let $\bigtriangleup ABC$ be isosceles with base BC. Then, $\angle B = \angle C$. Also, the median from vertex A, the bisector of $\angle A$, and the altitude from vertex A are all the same line. Prove this.

3) If two triangles have equal hypotenuses and an arm of one of the triangles equals an arm of the other, then the triangles are congruent. Prove.

4) An exterior angle of a triangle equals the sum of the two remote interior angles. Also, the sum of all three interior angles of a triangle is 180 degrees.

5) Find a formula for the interior angles of an n-gon.

6) Prove that the opposite sides of a parallelogram are equal.

7) In a quadrilateral ABCD, suppose that AB=CD and AD=BC. Then, prove that ABCD is a parallelogram.

8) In a quadrilateral ABCD, suppose that AB=CD and AB is parallel to CD. Then, prove that ABCD is a parallelogram.

9) Prove that a quadrilateral is a parallelogram iff its diagonals bisect each other.

10) Given a line segment BC, the locus of all points equidistant from B and C is the perpendicular bisector of the segment. Prove.

11) Corollary to problem 10 above: The diagonals of a rhombus are perpendicular. Prove.

12) Let AX be the bisector of $\angle A$ in $\triangle ABC$. Then, prove $\frac{BX}{XC} = \frac{AB}{AC}$. In other words, X divides BC into pieces proportional to the lengths of the nearer sides of the triangle. Prove.

13) Suppose that in $\triangle ABC$, the median from vertex A and the bisector of $\angle A$ are the same line. Show that $AB=AC$.

14) Prove that there is exactly one circle through any three given non collinear points.

15) An inscribed angle in a circle is equal in degrees to one half its subtended arc. Equivalently, the arc subtended by an inscribed angle is measured by twice the angle. Prove.

16) Corollary to above problem 15: Opposite angles of an inscribed quadrilateral are supplementary. Prove this.

17) Another corollary to above problem 15: The angle between two secants drawn to a circle from an exterior point is equal in degrees to half the difference of the two subtended arcs. Prove this.

18) A third corollary to above problem 15: The angle between two chords that intersect in the interior of a circle is equal in degrees to half the sum of the two subtended arcs. Prove this.

19) Theorem (Pythagoras): If a right triangle has arms of lengths a and b and its hypotenuse has length c, then $a^{2}+b^{2}=c^{2}$. Prove this.

20) Corollary to above theorem: Given a triangle ABC, the angle at vertex C is a right angle iff side AB is a diameter of the circumcircle. Prove this.

21) Theorem: The angle between a chord and the tangent at one of its endpoints is equal in degrees to half the subtended arc. Prove.

22) Corollary to problem 21: The angle between a secant and a tangent meeting at a point outside a circle is equal in degrees to half the difference of the subtended arcs.

23) Fix an integer, $n \geq 3$. Given a circle, how should n points on this circle be chosen so as to maximize the area of the corresponding n-gon?

24) Theorem: Given $\bigtriangleup ABC$ and $\bigtriangleup XYZ$, suppose that $\angle A = \angle X$ and $\angle B= \angle Y$. Then, prove that $\angle C = \angle Z$ and so $\bigtriangleup ABC \sim \bigtriangleup XYZ$. Prove this theorem.

25) Theorem: If $\bigtriangleup ABC \sim \bigtriangleup XYZ$, then the lengths of the corresponding sides of these two triangles are proportional. Prove.

26) The following lemma is important to prove the above theorem: Let U and V be points on sides AB and AC of $\bigtriangleup ABC$. Then, UV is parallel to BC if and only if $\frac{AU}{AB} = \frac{AV}{AC}$. You will have to prove this lemma as a part of the above proof.

27) Special case of above lemma: Let U and V be the midpoints of sides AB and AC, respectively in $\bigtriangleup ABC$. Then, UV is parallel to BC and $UV = \frac{1}{2}BC$.

28) Suppose that the sides of $\bigtriangleup ABC$ are proportional to the corresponding sides of $\bigtriangleup XYZ$. Then, $\bigtriangleup ABC \sim \bigtriangleup XYZ$.

29) Given $\bigtriangleup ABC$ and $\bigtriangleup XYZ$, assume that $\angle X = \angle A$ and that $\frac{XY}{AB} = \frac{XZ}{AC}$. Then, $\bigtriangleup ABC \sim \bigtriangleup XYZ$.

30) Consider a non-trivial plane geometry question now: Let P be a point outside of parallelogram ABCD and $\angle PAB = \angle PCB$. Prove that $\angle APD = \angle CPB$.

31) Given a circle and a point P not on the circle, choose an arbitrary line through P, meeting the circle at points X and Y. Then, the quantity $PX.PY$ depends only on the point P and is independent of the choice of the line through P.

32) You can given an alternative proof of Pythagoras’s theorem based on the following lemma: Suppose $\bigtriangleup ABC$ is a right triangle with hypotenuse AB and let CP be the altitude drawn to the hypotenuse. Then, $\bigtriangleup ACP \sim \bigtriangleup ABC \sim \bigtriangleup CBP$. Prove both the lemma and based on it produce an alternative proof of Pythagorean theorem.

33) Prove the following: The three perpendicular bisectors of the sides of a triangle are concurrent at the circumcenter of the triangle.

34) Prove the law of sines.

35) Let R and K denote the circumradius and area of $\bigtriangleup ABC$, respectively and let a, b and c denote the side lengths, as usual. Then, $4KR = abc$.

36) Theorem: The three medians of an arbitrary triangle are concurrent at a point that lies two thirds of the way along each median from the vertex of the triangle toward the midpoint of the opposite side.

37) Time to ponder: Prove: Suppose that in $\bigtriangleup ABC$, medians BY and CZ have equal lengths. Prove that $AB=AC$.

38) If the circumcenter and the centroid of a triangle coincide, then the triangle must be equilateral. Prove this fact.

39) Assume that $\bigtriangleup ABC$ is not equilateral and let G and O be its centroid and circumcentre respectively. Let H be the point on the Euler line GO that lies on the opposite side of G from O and such that $HG = 2GO$. Then, prove that all the three altitudes of $\bigtriangleup ABC$ pass through H.

40) Prove the following basic fact about pedal triangles: The pedal triangles of each of the four triangles determined by an orthic quadruple are all the same.

41) Prove the following theorem: Given any triangle, all of the following points lie on a common circle: the three feet of the altitudes, the three midpoints of the sides, and the three Euler points. Furthermore, each of the line segments joining an Euler point to the midpoint of the opposite side is a diameter of this circle.

42) Prove the following theorem and its corollary: Let R be the circumradius of triangle ABC. Then, the distance from each Euler point of $\bigtriangleup ABC$ to the midpoint of the opposite side is R, and the radius of the nine-point circle of $\bigtriangleup ABC$ is $R/2$. The corollary says: Suppose $\bigtriangleup ABC$ is not a right angled triangle and let H be its orthocentre. Then, $\bigtriangleup ABC$, $\bigtriangleup HBC$, $\bigtriangleup AHC$, and $\bigtriangleup ABH$ have equal circumradii.

43) Prove the law of cosines.

44) Prove Heron’s formula.

45) Express the circumradius R of $\bigtriangleup ABC$ in terms of the lengths of the sides.

46) Prove that the three angle bisectors of a triangle are concurrent at a point I, equidistant from the sides of the triangle. If we denote the by r the distance from I to each of the sides, then the circle of radius r centered at I is the unique circle inscribed in the given triangle. Note that in order to prove this, the following elementary lemma is required to be proved: The bisector of angle ABC is the locus of points P in the interior of the angle that are equidistant from the sides of the triangle.

47) Given a triangle with area K, semiperimeter s, and inradius r, prove that $rs=K$. Use this to express r in terms of the lengths of the sides of the triangle.

Please be aware that the above set of questions is almost like almost like a necessary set of pre-requisites for RMO geometry. You have to master the basics first.

Regards,

Nalin Pithwa.

# 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.

Amazon India link:

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.