Geometry and Trigonometry Toolkit — RMO | For the love of Maths: the Queen of all Sciences

Geometry:

The area of two triangles having equal bases (heights) are in the ratio of their heights (bases).
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.
Appolonius Theorem: If D is the mid-point of the side BC in a triangle ABC then $AB^{2}+AC^{2}=2(AD^{2}+BD^{2})$ .
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.
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.
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.
(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.
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.
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 $\angle{APB}$ .
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:

Compound and Multiple Angles:
- $\sin{(A \pm B)} = \sin{A}\cos{B} \pm \cos{A}\sin{B}$ ; $\cos{(A \pm B)} = \cos{A}\cos{B} \mp \sin{A}\sin{B}$ ; $\tan{(A \pm B)} = \frac{(\tan{A} \pm \tan{B})}{1 \mp \tan{A}\tan{B}}$ .
- $\sin{2A} = 2\sin{A}\cos{A} = \frac{2\tan{A}}{(1+\tan^{2}{A})}$ ; $\cos{2A}=\cos^{2}{A}-\sin^{2}{A}=2\cos^{2}{A}-1=1-2\sin^{2}{A}=\frac{1-\tan^{2}{A}}{1+\tan^{2}{A}}$ ; $\tan{2A}=\frac{2\tan{A}}{1-\tan^{2}{A}}$
- $\sin{3A}=3\sin{A}-4\sin^{3}{A}$ ; $\cos{3A}=4\cos^{3}{A}-3\cos{A}$ ; $\tan{3A}=\frac{(3\tan{A}-\tan^{3}{A})}{1-3\tan^{2}{A}}$ ;
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. $2\sin{A}\sin{B} = \cos{(A-B)} - \cos {(A+B)}$
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.
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:

http://www.amazon.in/Problem-Primer-Olympiad-2Ed-Pranesachar/dp/8172862059/ref=sr_1_1?s=books&ie=UTF8&qid=1492349245&sr=1-1&keywords=problem+primer+for+the+olympiad

Flipkart:

https://www.flipkart.com/problem-primer-olympiad-2nd/p/itmdyu5jyhwuwyvz?pid=9788172862053&srno=s_1_1&otracker=search&lid=LSTBOK97881728620535BELWH&qH=9abaf73d4a770fcd

More geometry later!

Nalin Pithwa

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