Tag Archives: Plane Geometry

Quick Review of Trigonometric Optimization Methods

Let us review together the four general methods we can use for triangular optimization.

I) Trigonometric Method: 

The essence of this method is the observation that the cosine of an angle is at most one and that it equals 1 only when the angle is zero. This fact is applied to the difference between two of the angles A, B and C, holding the third angle fixed, to show that unless those two angles are equal, the objective functionn can be increased (or decreased as the case may be). Consequently, at an optimal solution, these two angles must be equal. If the objective function is symmetric (as is the case, in almost all problems of triangular optimization), then every two of A, B and C must be equal to each other and hence, the triangle ABC must be equilateral.

This method is elementary and easy to apply. Even when the objective function is only partially symmetric, that is, symmetric in two but not in all the three variables, it can be applied to those two variables, holding the third variable fixed. Suppose, for example, that we want to maximize f(A,B,C)=\cos{A}+2\cos{B}+\cos{C}. This is symmetric in A and C. So, by the same reasoning as for maximizing \cos{A}+\cos{B}+\cos{C}, which at an optimal solution we must have A=C. Then, B=\pi-2A, which makes f effectively a function of just one variable, viz., \cos{A}+2\cos(\pi-2A)+\cos{A}, which equals 2\cos{A}-4\cos^{2}{A}+2. This can be maximized as a quadratic in \cos{A} either by completing the square or using calculus. The maximum occurs when A=\cos^{-1}{1/4}. Thus, the maximum value of \cos{A}+2\cos{B}+\cos{C} for a triangle ABC is 1/4. The method, of course, fails if the function is not even partially symmetric. This is not surprising. Basically, in a triangular optimization problem, we are dealing with a function f(A,B,C) of three variables. Because of the constraint A+B+C=\pi, any one of the variables can be expressed in terms of the other two. This effectively makes f a function of two variables. Optimization of functions of several variables requires advanced methods. It is only when f satisfies some other conditions such as partial symmetry that we can hope to reduce the number of variables further so that elementary methods can be applied.

II) Algebraic Method: 

The essence of this method is to reduce the optimization problem to some inequality using suitable trigonometric formulae or identities. The inequality is then established using some standard inequality such as the AM-GM-HM inequality, or Jensen’s inequality, or sometimes, by doing some more basic work. The fundamental ideas are very simple, viz., (a) the square of any real number is non-negative and is zero only when that number is zero, and (b) the sum of two or more non-negative numbers is non-negative and vanishes if and only if each of the term is zero. When this method works, it works elegantly. But it is not always easy to come up with the right algebraic manipulations. Sometimes, certain simplifying substitutions have to be used. Still, it is an elementary method and deserves to be tried.

III) Jensen’s inequality:

This is a relatively advanced method. It is directly applicable when the objective function is, or can be recast, in a certain form, viz., h(A)+h(B)+h(C), where h is a function of one variable whose second derivative maintains the same sign over a suitable interval. But, even when h fails to do so, the method can sometimes be applied with a suitable conversion of the problem.

IV) Lagrange’s Multipliers:

This is a highly advanced method based on the calculus of functions of several variables. It is applicable to all types of objective functions, not just those that are symmetric or partially symmetric. When applied to triangular optimization problems with symmetric objective functions, the optimal solution is either degenerate or an equilateral triangle.

Naturally, for a particular given problem, some of these methods may work better than others. The method of Lagrange’s multipliers is the surest but the most mechanical of all the four. The algebraic method is artistic and sometimes gives the answer very fast. Jensen’s inequality also works fast once you are able to cast the objective function in a certain form. Such a recasting may involve some ingenuity sometimes. The trouble is that both these methods work only in the case of an  optimization problem where the objective function is symmetric. And, in such cases, the method of Lagrange’s multipliers makes a mincemeat of the problem. From an examination point of view, this is a boon if a question about triangular optimization is asked in a “fill in the blanks” or “multiple choice” form, where you don’t have to show any reasoning. if the objective function is symmetric, then the optimal solution is either degenerate or an equilateral triangle. But, degenerate triangles are often excluded from the very definition of a triangle because of the requirement that the three vertices of a triangle must be distinct and non-collinear and, in any case, such absurdities are unlikely to be asked in an examination! So, it is a safe bet to simply assume that the optimal solution is an equilateral triangle and proceed with further work (namely, calculating the value of the objective function for an equilateral triangle). This saves you a lot of time.

More later,

Nalin Pithwa

 

Proof of SAS Congruency Test of two triangles

SAS Test math blogTheorem: If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, then the triangles are equal in all respects.

Construction: Let ABC, DEF be two triangles in which AB=DE and

AC=DF and the included angle \angle BAC is equal to included angle \angle EDF.

It is required to prove that the

\Delta ABC = \Delta DEF in all respects.

Proof: Apply the \Delta ABC to the \Delta DEF so that the point A falls on the point D; and the side AB along the side DE. Then, because AB=DE so the point B must coincide with the point E. And, because AB falls along DE, and the

\angle BAC=\angle EDF, so AC must fall along DF. And, because

AC=DF, the point C must coincide with the point F. Then, since B coincides with E, and C with F, hence, the side BC must coincide with the side EF. Hence, the \Delta ABC coincides with the \Delta DEF, and is therefore equal to it in all respects. QED.

More later…please post your questions, comments, and I will gladly answer them asap

Nalin Pithwa

Congruency Tests of two triangles — for PreRMO, RMO, IITJEE Foundation Math

Uptil now, we have discussed quite a bit of algebra, number theory and calculus for RMO, IITJEE Main and Advanced Mathematics. Let us switch back to plane geometry for a while.

In school, we all learn about tests of congruency of two triangles. But, have you wondered what is the exact meaning of “two plane figures are congruent”? Try to Google and find the answer for yourself. You can also post your answers as comments to this blog article.

The reasons I discuss these are two fold: (1) do you know the proofs of these congruency tests? In schools, students are just taught the meaning of these and how to apply them. Well, I will give you proofs of these in my blog at some later time. (2) There are some nuances to  these tests; when they cannot be applied, or rather, which *tests* fail to apply.

Here we go…

Two triangles are equal in all respects when the following three parts in each are severally equal:

1)Two sides, and the included angle. (the SAS Test for Congruency of 2 triangles).

2) The three sides. (the SSS Test for congruency of two triangles.

3) Two angles and one side, the side given in one triangle CORRESPONDING to that given in the other (ASA and AAS Test for Congruency of two triangles).

The two triangles are not, however, necessarily equal in all respects, when any three parts of one are equal to the corresponding parts of the other.

For example:

1) When the three angles of one are are equal to  the three angles of the  other, each to each, the fig 1 (attached, please download) shows  that the triangles need not be equal in all respects. (If you recall, such triangles are called similar, not  congruent because congruent means exactly same; two twins are similar, but not same).

2) When two sides and one angle in one are equal to two sides and one angle of the other, the given angles being opposite to equal sides, the  fig 2 (attached, please download) shows that the triangles need not be equal in all respects.

For, if $AB=DE$ and $AC=DF$ and the $\angle ABC = \angle DEF$, it will be seen that the shorter of the given sides in the triangle DEF may lie in either of  the positions DF or DF’.

Important Note: From these data, it may be shown that the angle opposite to the equal sides AB, DE are either equal (as for instance, the $\angle ACB$ and

$\angle DF’E$) or supplementary as (the $\angle ACB and \angle DFE$); and, that in the former case the triangles are equal in all respects. This is called the ambiguous case in the congruence of triangles. 

Note: ambiguous means that which has some uncertainty. In other words, there could be more than one possibilities.

If the given angles at B and E are right angles, the ambiguity disappears.

In  the next few blog articles, we will discuss the proofs of congruency tests of two triangles.

More later,

Nalin PIthwablogfigcongruencetests