A Triangle With a 45 Degrees Angle in Square : A very hard problem from Geometry

Let P and Q be the points where AN and AM intersect the diagonal BD, respectively. It is noted that ∠QAN=∠QDN=45∘.Thus, AQND is a cyclic quadrilateral, therefore implying that ∠AQN=90∘.

This in turn implies that ΔAQN is a right-isoceles triangle with AN as the hypotenuse. Thus, AQ=AN2–√.

The same argument can be employed to show that AP=AM2–√.

Now, since triangles APQ and AMN share a common vertex angle A, we have, Area(ΔAPQ)Area(ΔAMN)=AQ⋅APAM⋅AN. However, from the conclusions in the previous two paragraphs, we have AQAN=APAM=12–√.

This therefore implies that Area(ΔAPQ)Area(ΔAMN)=12, or equivalently, Area(ΔAPQ)Area(MNPQ)=1, as claimed by the problem.

The problem, due to V. Proizvolov, appeared in Kvant - a popular Russian magazine - (#1, 2004, M1895), with a solution in a later issue (#4, 2004).

Class IX West Bengal Board Ganit Prakash Chapter 20 Solutions- Co-ordinate Geometry

Class IX West Bengal Board Ganit Prakash Chapter 20 Solutions - Coordinate Geometry

In this pdf you will get all the solved problem from the exercise 20. Before going through the solutions you are advised to first try the problem yourself. If you get stuck, seek help from the solutions. Don't copy blindly. If you have any query, comment below.

Can you solve the problem involving Intersection Chords Theorem?

Two chords AB and CD of a circle intersects perpendicularly at the point P. The length of the segments AP, PB, CP and PD are given. Can you calculate the radius of the circle? Watch the full video for an interesting solution with the help of intersecting chord theorem? Learn something new and different.

Can you solve the system of non-linear equation?

A system of linear equations can be solved easily by many standard methods. Even in high school mathematics courses, students are taught Cramer's rule and the method of Matrix to solve a given set of consistent linear equations up to three variables. More advance methods are taught in the course of linear algebra. But for a given system of non-linear equations, there is no specific method. Different ideas need to be applied across various parts of mathematics to get the solution. Nonetheless using school level mathematics, many interesting problems can be solved. Here we present once such problem, where we have to solve a system of non-linear equation. x^2 -  yz = 3, y^2 - zx = 4 and z^2 - xy = 5 is the system here, we will solve it using elementary ideas. Knowledge of arithmetic progression is essential.

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Can You Solve The 5 Circles inside the square Problem?

Five circles with the same radius are placed inside a square, whose sides are not known. The radius of the circle is given. The problem is to find the side of the square. Four circles are placed inside the square at the four corners in such a way that the sides of the square are tangents to the circle. The fifth circle is place between the four circles, such that it touches the four circles externally.

Watch the video for the solution

Solved Problems : Circles for Class X

Solved Problems: Trigonometry - Angles and Sides of a Triangle.

To get the pdf file visit the page: http://kolkatamaths.yolasite.com/xi-xii.php
File name: Angles and Sides of a Triangle_Loney

Trigonometry: Elimination Problem

Solving a system of in-equations: A problem form High School

In this video, I have solve a system of in-equations, using elementary ideas of high school mathematics. Using the fact that, square of any real number is always greater than equal to zero, we can at the same time frame and solve unique problems. Watch the full video, to learn.

Less calculations, more thought: Problem form Definite Integral

A very interesting problem from, definite integrals. This problem is slightly different form the usual problems of definite integral, as the function given here is not the usual polynomial or trigonometric functions. Watch the full video for the solution and learn something new and different.

Algebra Problem: Can you Solve it?

Using elementary identity from middle school, his problem can be solved. But it's really challenging and tricky if you don't hit the right idea. Check out the video to learn something new and interesting.

Let x,y,z be distinct real
numbers.

Prove that

∛(x-y)+∛(y-z)+∛(z-x)  ≠0

You should be able to solve this! Minimum value Problem

A problem of minimisation. Given an expression, for which we have to find the minimum value. A very interesting problem, with simple properties of logarithmic functions to be used. Watch the full video to see the solution and learn something new and different.

A very interesting problem from Indian Statistical Institute: Geometry

This problem, requires some knowledge of a regular polygon. Only the elementary properties is required. If you don't know anything about regular polygons, just google it! Lot of thinking is required to solve this problem. A very good problem where the calculations are minimal but the logical part is required throughout. Check the full video for the solution and share your views about the problem.

Problem form American Invitational Mathematics Examination: Can you solve it?

A very beautiful problem with elegant solution. Suitable for students preparing for IIT and other entrance exams. Even for school students this problem is suited who wants to learn something different. As the problem doesn't require more than knowledge of multiplication of variables and solving simultaneous linear equations, a 10th grade students can also understand it!

An interesting property of cyclic quadrilateral.

In a cyclic quadrilateral the product of its diagonals is equal to the sum of the products of it's opposite sides. Watch the full video for the proof. 

Definite Integral : Problem with a twist!

Sum of squares of 5 consecutive natural numbers!

Here's a difficult problem of proving that a given expression is not a perfect square. A direct approach will be a nightmare ( even not sure it can be proved) but use of a simple property of perfect squares will ease the problem. We know that any perfect squares leaves a remainder either 1 or 0 when being divided by 3 or 4. This simple result would be used to solve the problem.

Number of solutions of the given equation

Here the equation to be solved is very simple |2x-[x]|=4. But unfortunately the traditional methods of solving will not apply! You have solved equations by methods like factorisation, vanishing method, hit and trial.... But in many equations these methods don't work or requires huge computation and guess work to solve it. Here is an equation which can be solved from the simple fact that if one side of an equation is an integer the we look to the other side only and see for what values its also an integer. This simple but powerful method can be used in many problems

A very hard integration problem! Can you solve it? ( Practice for IIT-JEE)

A very hard integration problem based on the method of substitution. It took almost an hour to figure this out! Can you solve it on your own?! This problem is well suited for students preparing for IIT-JEE.

Inequalities

Show that $|x|+|y| \leq |x+y|+|x-y| \quad and \quad \frac{|x+y|}{1+|x+y|} \leq \frac{|x|}{1+|x|} + \frac{|y|}{1+|y|} \quad for \quad x,y \in \mathbb{R} $ $$$$ We have $\quad|x+y+x-y| \leq |x+y|+|x-y| $ $$$$ $ \implies 2|x| \leq |x+y|+|x-y| \dots A$ $$$$ Now, $|x+y+y-x| \leq |x+y|+|y-x|$ $$$$ $ \implies 2|y| \leq |x+y|+|-1||x-y| $ $$$$ $\implies 2|y| \leq |x+y|+|x-y| \dots B$ $$$$ Adding $A$ and $B$ we get $ |x|+|y| \leq |x+y|+|x-y|$ The second inequality holds $$$$ $iff$ $\frac{(1+|x|)(1+|y|)|x+y|}{1+|x+y|} \leq |x|(1+|y|)+|y|(1+|x|)$ $$$$ $ iff \quad (1+|x|)(1+|y|)|x+y| \leq (|x|+|y|+2|xy|)(1+|x+y|)$ $$$$ (multiply and cancel out terms) $$$$ $iff \quad |x+y| \leq |x|+|y| +|xy|(2+|x+y|)$ $$$$ Now for any real $x$ and $y$, $|xy|(2+|x+y|) \geq 0$ and we know that $|x+y| \leq |x|+|y|$ $$$$ Thus the last inequality holds, which in turn proves the original ineqality.

Solved Trigonometry Problems: 10th Grade

Here is a list of few good problems at the 10th grade. Get the full solution below and if you want
more comment or mail us at maths.programming@gmail.com!


Download here

Circle: Geometry

Given below is a fine problem based on simple properties of circles ( tangent and chord ). Try yourself before looking at the solution!

Here is the solution!

Solved Problems : Logarithm

A collection of solved problems on logarithm meant for secondary students ( 9th and 10th grade). The collection covers almost all types of problems at the level mentioned.



Download the file here: Solved Logarithm Problems 

Solved Integration Problems for 10+2 Level - SN Dey

Hundreds of solved problems at 10+2 level or the final year at school from the topic of integration. The aim of the document to help students from different boards (ISC, CBSE and other State Boards) studying at Higher Secondary level to get access to wide variety of problems, that too solved! Most of the problems are from the books of SN Dey, which is the most sought after book in WBCHSE.

This should not be substituted for classroom teaching neither this documents aims to teach the theories of differentiation. Students are requested to go through the theory and the rules before looking at the problems. Any error in the document can be reported at: prime.maths@hotmail.com

Integration, Methods of Substitution, Integration by Parts, Partial Fractions, Special Integrations, Trigonometric Substitutions.

If you have any query, don't forget to comment below.

A  collection of completely solved special integrals for various entrance exams ( IIT and Indian Statistical Institute ) and boards exam (CBSE,ISC and other State Boards) at 10+2 level. 

Click below the links to download the files:

File 1 Introduction ( Basic Problems )

File 2 Special Integrals 

File 3 Method of Substitution- Set I

File 4 Method of Substitution- Set II

File 5 Method of Substitution- Set II

File 6 Problems based on Standard Integrals- Set I

File 7 Problems based on Standard Integrals- Set II 

File 8 Problems based on Standard Integrals- Set III

File 9 Problems based on Integration by Parts - Set I

File 10 Problems based on Integration by Parts - Set II

More solved problems to follow. Keep visiting this space.

Special Integration

Integration : A harder Problem

 Most of the studenst will fail to solve this particular integration problem. It is trickier but once you hit the right idea, you will be able to solve the integration problem easily.

A problem on inequality

Using simple formula to prove a strong inequality.

Pigeonhole Principle

The numbers 1 to 20 are placed in any order around a circle. Prove that the sum of some 3 consecutive numbers must be at least 32!

This problem uses the alternate form of pigeon hole principle which is as follows:
 If the average of n positive numbers is t, then at least one of the numbers is greater than or equal to t. Further, at least one of the numbers is less than or equal to t.

The proof is very simple, assume the contradiction and proceed!

#Solution https://youtu.be/GLQg6cSAbms

Sum of first 'n, natural numbers

A simple way to calculate the sum of first 'n' natural numbers witout the use of calculator. Infact the same procedure is used to calculate the sum of n terms of any A.P series. See it and try to obtain the formula yourself

Inequality

A challenging problem based on the inequality that square of a real number is always greater than equal to zero. Learn the trick and prepare yourself for more challenging problems based on the same ideology.

Divisibility of 2222^5555+5555^2222 by 7

For any prime p, square root of p is irrational.

Complex Numbers in Solving Trigonometrical problems

Show that $ \cos \frac{\pi}{11}+\cos \frac{3\pi}{11}+\cos \frac{5\pi}{11}+\cos 7\frac{\pi}{11}+\cos \frac{9\pi}{11}= \frac{1}{2}$ $$$$ I have used complex numbers to solve the problems. Standard trigonometric methods will be very much difficult to apply and will involve lots of calculation!!! Also it is important to link the different topics and see the beauty of mathematics! Important problem for students in 10+2 level $$$$ Also the idea used to solve the problem can be used to calculate the value of $ \sum_{i=1}^{\frac{n-1}{2}} \cos \frac{(2i-1) \pi}{n}$ where $n$ is an odd positive integer. $$$$ See the video below for the explanation. $$$$

Equation Solving

Find all real number(s) $x$ such that $2^x+3^x+6^x-4^x-9^x=1 $ $$$$ This problem can be solved by transforming the equation where it can be written as sum of squares of number equals zero. From there the result will follow. See the video below for the explanation. $$$$

Sophie Germain Identity

Here is a problem of primality testing of a number. As you can see a direct computation will not yield an effective result and will take a much longer time! So, the question arises how to solve the problem in the right way. Here comes the roll of simple and elegant identity known as Sophie Germain Identity. Watch the video below for the solution

For solution of the problem above watch the video below.

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Algebra

Let $a, b$ and $c$ be such that $a+b+c=0$ and \( P =\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\) is defined. What is the value of $P$? $$$$ $Solution: $ \( 2a^2+bc = a^2 + a^2 -(a+c)c\), since $b=-(a+c)$. $$$$ \( \implies 2a^2+bc = a^2 + a^2 -ac-c^2 = (a-c)(a+c)+a(a-c) = (a-c)(2a+c)=(a-c)(2a-a-b)=(a-c)(a-b)\) $$$$ Therefore \( \frac{a^2}{2a^2+bc} = \frac{a^2}{(a-c)(a-b)}\) $$$$ Similarly, \( \frac{b^2}{2b^2+ac} = \frac{b^2}{(b-a)(b-c)} \) and \( \frac{c^2}{2c^2+ab} = \frac{c^2}{(c-a)(c-b)} \) $$$$ Therefore \( P = \frac{a^2}{(a-c)(a-b)} +\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-a)(c-b)} = \frac{1}{a-b} \big( \frac{a^2}{a-c} - \frac{b^2}{b-c} \big)+\frac{c^2}{(c-a)(c-b)}\) $$$$ \( \implies P = \frac{ab-ca-cb}{(c-a)(c-b)}+ \frac{c^2}{(c-a)(c-b)} = \frac{ab-ca-cb+c^2}{(c-a)(c-b)} = \frac{(c-a)(c-b)}{(c-a)(c-b)} = 1\)$$$$

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Calculus

If \( A = \int_{0}^{\pi} \frac{cos x}{(x+2)^2}dx\), then show that \( \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{(x+1)}dx = \frac{1}{2} \big( \frac{1}{2} +\frac{1}{\pi+2} - A\big)\). $$$$ Solution: \( A = \cos x \int_{0}^{\pi} \frac{dx}{(x+2)^2} - \int_{0}^{\pi} \bigg( \frac{d}{dx} \cos x \int \frac{dx}{(x+2)^2} \bigg) dx \) $$$$ \( \implies A = \frac{-\cos x}{x+2}\mid_{0}^{\pi} - \int_{0}^{\pi} \frac{\sin x}{(x+2)} dx \) $$$$ \( \implies A = \frac{1}{\pi +2} + \frac{1}{2} - \int_{0}^{\pi} \frac{2 \sin \frac{x}{2} \cos \frac{x}{2} }{2(1+\frac{x}{2})} dx \) $$$$ Putting $\frac{x}{2}=t$ in the last integral, we have $dx=2dt$ and $t$ varies from $0$ to $\frac{\pi}{2}$ $$$$ Therefore, \( A = \frac{1}{\pi +2} + \frac{1}{2} - 2 \int_{0}^{\frac{\pi}{2}} \frac{\sin t \cos t}{(1+t)}dt \)$$$$ \( \implies \int_{0}^{\frac{\pi}{2}} \frac{\sin t \cos t}{(1+t)}dt = \frac{1}{2} \bigg( \frac{1}{\pi +2} + \frac{1}{2} -A \bigg) \) $$$$ \( i.e., \) \( \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{(1+x)}dx = \frac{1}{2} \bigg( \frac{1}{\pi +2} + \frac{1}{2} -A \bigg) \) $$$$

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Calculus

Evaluate \( lim_{n \rightarrow \infty} \bigg(\prod_{r=1}^{r=n} (1+\frac{2r-1}{2n}) \bigg)^{\frac{1}{2n}} \) $$$$ Let \( A = \bigg(\prod_{r=1}^{r=n} (1+\frac{2r-1}{2n}) \bigg)^{\frac{1}{2n}} \) $$$$ \( \implies A = \frac{\bigg(\prod_{r=1}^{r=2n} (1+\frac{r}{2n}) \bigg)^{\frac{1}{2n}}}{\bigg(\prod_{r=1}^{r=n} (1+\frac{2r}{2n}) \bigg)^{\frac{1}{2n}}} \) $$$$ \( \implies ln A = \frac{1}{2n} \sum_{r=1}^{r=2n} ln ( 1+\frac{r}{2n})- \frac{1}{2n} \sum_{r=1}^{r=n} ln ( 1+\frac{r}{n})\) $$$$ \( \implies lim_{n \rightarrow \infty} ln A = \frac{1}{2} lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{r=2n} ln ( 1+\frac{r}{2n}) - \frac{1}{2} lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{r=n} ln ( 1+\frac{r}{n}) \) $$$$ \( \implies lim_{n \rightarrow \infty} ln A = \frac{1}{2} \int_{0}^{2} ln (1+ \frac{x}{2}) dx - \frac{1}{2} \int_{0}^{1} ln (1+ x ) dx \) $$$$ \( \implies lim_{n \rightarrow \infty} ln A = ln 4 - 1 - ( ln 2 - \frac{1}{2}) = ln 2 - \frac{1}{2}\)$$$$ \( \implies lim_{n \rightarrow \infty} A = e^{ln 2 - \frac{1}{2}} = \frac{2}{\sqrt{e}} \) $$$$ \( \implies lim_{n \rightarrow \infty} \bigg(\prod_{r=1}^{r=n} (1+\frac{2r-1}{2n}) \bigg)^{\frac{1}{2n}} = \frac{2}{\sqrt{e}} \) $$$$

ISC Mathematics Paper 2016 Solved

Get the solved ISC Mathematics Question paper for the year 2016. The solutions are short need not to be considered as the way of answering in board exams. Click Below 
ISC MATHEMATICS PAPER 2016


ISC MATHEMATICS PAPER 2015

Solved CBSE Mathematics Question paper 2015. 

JEE MAIN 2016 MATHEMATICS SOLVED PAPER

JEE MAIN 2015 MATHEMATICS SOLVED PAPER

JEE MAIN 2016 Mathematics

Solved Paper of JEE MAIN 2016 Mathematics.

Download Link : Click Here

Binomial Series

Find the sum of the series \( \sum_{r=0}^{n} (-1)^r \binom{n}{r} \big( \frac{1}{2^r}+\frac{3^r}{2^{2r}}+\frac{7^r}{2^{3r}}+ \dots \) to m terms \( \big) \) $$$$ Given series is equal to \( \sum_{r=0}^{n} (-1)^r \binom{n}{r} \sum_{k=1}^{m} \frac{(2^k -1)^r}{2^{kr}} \) $$$$ \( = \sum_{r=0}^{n} \sum_{k=1}^{m} (-1)^r \binom{n}{r} \frac{(2^k -1)^r}{2^{kr}} \) $$$$ \( = \sum_{r=0}^{n} \sum_{k=1}^{m} (-1)^r \binom{n}{r} \lambda^r \) where $\lambda = \frac{(2^k -1)}{2^{k}}$ $$$$ \( = \sum_{k=1}^{m} \sum_{r=0}^{n} (-1)^r \binom{n}{r} \lambda^r \) Interchanging the summation $$$$ \( = \sum_{k=1}^{m} (1- \lambda)^n \), Substituting the value of $\lambda$ we have, $$$$ \( = \sum_{k=1}^{m} \frac{1}{2^{nk}} = \frac{1}{2^n} \frac{\bigg(1- \big(\frac{1}{2^n}\big)^m \bigg)}{1 - \frac{1}{2^n} } \) $$$$ \( = \frac{1}{2^n} \frac{2^n(2^{nm}-1)}{2^{nm}(2^n-1)} = \frac{2^{nm}-1}{2^{nm}(2^n-1)} \)

Integration

Evaluate \(\int \frac{dx}{e^x \big(1+e^{2008x}\big)^{\frac{2007}{2008}}} \) $$$$ \(\int \frac{dx}{e^x \big(1+e^{2008x}\big)^{\frac{2007}{2008}}} = \int \frac{dx}{e^x \big(e^{2008x}(1+e^{-2008x})\big)^{\frac{2007}{2008}}}\) $$$$ \(=\int \frac{dx}{e^x e^{2007x}\big(1+e^{-2008x}\big)^{\frac{2007}{2008}}} = \int \frac{e^{-2008x}dx}{\big(1+e^{-2008x}\big)^{\frac{2007}{2008}}} \) \(= \frac{-1}{2008}\int \frac{d(1+e^{-2008x})}{\big(1+e^{-2008x}\big)^{\frac{2007}{2008}}} = \frac{\frac{-1}{2008}\big(1+e^{-2008x}\big)^{-\frac{2007}{2008}+1}}{-\frac{2007}{2008}+1} + c \)