A quadratic and trigonometry combo question: RMO and IITJEE maths coaching


Given that \tan {A} and \tan {B} are the roots of the quadratic equation x^{2}+px+q=0, find the value of

\sin^{2}{(A+B)}+ p \sin{(A+B)}\cos{(A+B)} + q\cos^{2}{(A+B)}


Let \alpha=\tan{A} and \beta=\tan{B} be the two roots of the given quadratic equation: x^{2}+px+q=0

By Viete’s relations between roots and coefficients:

\alpha+\beta=\tan{A}+\tan{B}=-p and \alpha \beta = \tan{A}\tan{B}=q but we also know that \tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\frac{-p}{1-q}=\frac{p}{q-1}

Now, let us call E=\sin^{2}{(A+B)}+p\sin{(A+B)\cos{(A+B)}}+\cos^{2}{(A+B)} which in turn is same as


We have already determined \tan{(A+B)} in terms of p and q above.

Now, again note that \sin^{2}{\theta}+\cos^{2}{\theta}=1 which in turn gives us that \tan^{2}{\theta}+1=\sec^{2}{\theta} so we get:

\sec^{2}{(A+B)}=1+\tan^{2}{(A+B)}=1+\frac{p^{2}}{(q-1)^{2}}=\frac{p^{2}+(q-1)^{2}}{(q-1)^{2}} so that


Hence, the given expression E becomes:

(\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}})(\frac{p^{2}}{(q-1)^{2}}+\frac{p^{2}}{q-1}+q), which is the desired solution.

πŸ™‚ πŸ™‚ πŸ™‚

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.


Refer Wikipedia πŸ™‚ πŸ™‚ πŸ™‚



Nalin Pithwa.


RMO Training: a question from Nordic Mathematical Contest: 1988


Two spheres with the same centre have radii r and R, where r<R. From the surface of the bigger sphere we will try to select three points A, B and C such that all sides of the triangle ABC coincide with the surface of the smaller sphere. Prove that this selection is possible if and only if R<2r.


Assume A, B and C lie on the surface \Gamma of a sphere of a radius R and centre O, and AB, BC, and CA touch the surface \gamma of a sphere of radius r and centre O. The circumscribed and inscribed circles of ABC then are the intersections of the plane ABC with \Gamma and \gamma, respectively. The centres of these circles both are the foot D of the perpendicular dropped from O to the plane ABC. This point lies both on the angle bisectors of the triangle ABC and on the perpendicular bisectors of its sides. So, these lines are the same, which means that the triangle ABC is equilateral, and the centre of the circles is the common point of intersection of the medians of ABC. This again implies that the radii of the two circles are 2r_{1} and r_{1} for some real number r_{1}. Let OD=d. Then, 2r_{1}=\sqrt{R^{2}-d^{2}} and r_{1}=\sqrt{r^{2}-d^{2}}. Squaring, we get R^{2}-d^{2}=4r^{2}-4d^{2}, 4r^{2}-R^{2}=3d^{2} \geq 0, and 2r\geq R.

On the other hand, assume 2r \geq R. Consider a plane at the distance d=\sqrt{\frac{4r^{2}-R^{2}}{3}} from the common centre of the two spheres. The plane cuts the surfaces of the spheres along concentric circles of radii

r_{1}=\sqrt{\frac{R^{2}-r^{2}}{3}} and R_{1}=2\sqrt{\frac{R^{2}-r^{2}}{3}}.

The points A, B, and C can now be chosen on the latter circle in such a way that ABC is equilateral.

Reference: Nordic Mathematical Contest, 1987-2009.


Nalin Pithwa.

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

Well, the solutions already exist ! (pun intended! πŸ™‚ πŸ™‚ :-))

You may note that putting one of Β 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!

πŸ™‚ πŸ™‚ πŸ™‚

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


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.Β 

Geometry and Trigonometry Toolkit — RMO


  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.


  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}} Β = 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:Β To the student readers: from Ref: Problem Primer for the Olympiad by C. R. Pranesachar, B. J. Venkatachala, C. S. Yogananda:Β 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.Β 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:




More geometry later!

Nalin Pithwa

Heights and Distances — II — problems for RMO and IITJEE maths

This is one of the prime utilities of trigonometry —- to calculate heights and distances, called as surveying in civil engineering.

1.Β From the extremities of a horizontal base line AB, whose length is 1 km, the bearings of the foot C of a tower are observed and it is found that \angle {CAB} is 56 degrees and 23 minutes and \angle{CBA} is 47 degrees and 15 minutes, and the elevation of the tower from A is 9 degrees and 25 minutes, find the height of the tower.

2.Β A man in a balloon observes that the angle of depression of an object on the ground bearing due north is 33 degrees; the balloon drifts 3 km due west and the angle of depression is now found to be 21 degrees. Find the height of the balloon.

3. A tower PN stands on level ground. A base AB is measured at right angles to AN, the points A, B, and N being in the same horizontal plane, and the angles PAN and PBN are found to be \alpha and \beta respectively. Prove that the height of the tower is

AB\frac{\sin {\alpha}\sin{\beta}}{\sqrt{\sin{(\alpha-\beta)}\sin{(\alpha+\beta)}}}

If AB is 100m, \alpha = 70 degrees and \beta = 50 degrees, calculate the height.

4. At each end of a horizontal base of length 2a, it is found that the angular height of a certain peak is \theta and that at the middle point it is \phi. Prove that the vertical height of the peak is

\frac{a \sin {\theta}\sin{\phi}}{\sqrt{\sin{(\phi+\theta)}\sin{(\phi-\theta)}}}

5. To find the distance from A to P, a distance AB of 1 km. is measured for a convenient direction. At A the angle PAB is found to be 41 degrees 18 min and at B, the angle PBA is found to be 114 degrees and 38 min. What is the required distance to the nearest metre?

More later,

Nalin Pithwa