Awesome inspiration for Mathematics !

The Early History of Calculus Problems, II: AMS feature column

The Early History of Calculus Problems: AMS Feature column

Vladimir Voevodsky: Obit: The Washington Post



Vladimir Voevodsky: Obit: The New York Times

Fun with Math websites: MAA

cheers to MAA! 🙂

Nalin Pithwa.

Cartesian System, Straight Lines: IITJEE Mains: Problem Solving Skills II

I have a collection of some “random”, yet what I call ‘beautiful” questions in Co-ordinate Geometry. I hope kids preparing for IITJEE Mains or KVPY or ISI Entrance Examination will also like them.

Problem 1:

Given n straight lines and a fixed point O, a straight line is drawn through O meeting lines in the points R_{1}, R_{2}, R_{3}, \ldots, R_{n} and on it a point R is taken such that \frac{n}{OR} = \frac{1}{OR_{1}} + \frac{1}{OR_{2}} + \frac{1}{OR_{3}} + \ldots + \frac{1}{OR_{n}}

Show that the locus of R is a straight line.

Solution 1:

Let equations of the given lines be a_{i}x+b_{i}y+c_{i}=0, i=1,2,\ldots, n, and the point O be the origin (0,0).

Then, the equation of the line through O can be written as \frac{x}{\cos{\theta}} = \frac{y}{\sin{\theta}} = r where \theta is the angle made by the line with the positive direction of x-axis and r is the distance of any point on the line from the origin O.

Let r, r_{1}, r_{2}, \ldots, r_{n} be the distances of the points R, R_{1}, R_{2}, \ldots, R_{n} from O which in turn \Longrightarrow OR=r and OR_{i}=r_{i}, where i=1,2,3 \ldots n.

Then, coordinates of R are (r\cos{\theta}, r\sin{\theta}) and of R_{i} are (r_{i}\cos{\theta},r_{i}\sin{\theta}) where i=1,2,3, \ldots, n.

Since R_{i} lies on a_{i}x+b_{i}y+c_{i}=0, we can say a_{i}r_{i}\cos{\theta}+b_{i}r_{i}\sin{\theta}+c_{i}=0 for i=1,2,3, \ldots, n

\Longrightarrow -\frac{a_{i}}{c_{i}}\cos{\theta} - \frac{b_{i}}{c_{i}}\sin{\theta} = \frac{1}{r_{i}}, for i=1,2,3, \ldots, n

\Longrightarrow \sum_{i=1}^{n}\frac{1}{r_{i}}=-(\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})\cos{\theta}-(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})\sin{\theta}

\frac{n}{r}=-(\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})\cos{\theta}-(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})\sin{\theta} …as given…

\Longrightarrow (\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})r\cos{\theta}+(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})r\sin{\theta} + n=0

Hence, the locus of R is (\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})x+(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})y+n=0 which is a straight line.

Problem 2:

Determine all values of \alpha for which the point (\alpha,\alpha^{2}) lies inside the triangle formed by the lines 2x+3y-1=0, x+2y-3=0, 5x-6y-1=0.

Solution 2:

Solving equations of the lines two at a time, we get the vertices of the given triangle as: A(-7,5), B(1/3,1/9) and C(5/4, 7/8).

So, AB is the line 2x+3y-1=0, AC is the line x+2y-3=0 and BC is the line 5x-6y-1=0

Let P(\alpha,\alpha^{2}) be a point inside the triangle ABC. (please do draw it on a sheet of paper, if u want to understand this solution further.) Since A and P lie on the same side of the line 5x-6y-1=0, both 5(-7)-6(5)-1 and 5\alpha-6\alpha^{2}-1 must have the same sign.

\Longrightarrow 5\alpha-6\alpha^{2}-1<0 or 6\alpha^{2}-5\alpha+1>0 which in turn \Longrightarrow (3\alpha-1)(2\alpha-1)>0 which in turn \Longrightarrow either \alpha<1/3 or \alpha>1/2….call this relation I.

Again, since B and P lie on the same side of the line x+2y-3=0, (1/3)+(2/9)-3 and \alpha+2\alpha^{2}-3 have the same sign.

\Longrightarrow 2\alpha^{2}+\alpha-3<0 and \Longrightarrow (2\alpha+3)(\alpha-1)<0, that is, -3/2 <\alpha <1…call this relation II.

Lastly, since C and P lie on the same side of the line 2x+3y-1=0, we have 2 \times (5/4) + 3 \times (7/8) -1 and 2\alpha+3\alpha^{2}-1 have the same sign.

\Longrightarrow 3\alpha^{2}+2\alpha-1>0 that is (3\alpha-1)(\alpha+1)>0

\alpha<-1 or \alpha>1/3….call this relation III.

Now, relations I, II and III hold simultaneously if -3/2 < \alpha <-1 or 1/2<\alpha<1.

Problem 3:

A variable straight line of slope 4 intersects the hyperbola xy=1 at two points. Find the locus of the point which divides the line segment between these two points in the ratio 1:2.

Solution 3:

Let equation of the line be y=4x+c where c is a parameter. It intersects the hyperbola xy=1 at two points, for which x(4x+c)=1, that is, \Longrightarrow 4x^{2}+cx-1=0.

Let x_{1} and x_{2} be the roots of the equation. Then, x_{1}+x_{2}=-c/4 and x_{1}x_{2}=-1/4. If A and B are the points of intersection of the line and the hyperbola, then the coordinates of A are (x_{1}, \frac{1}{x_{1}}) and that of B are (x_{2}, \frac{1}{x_{2}}).

Let R(h,k) be the point which divides AB in the ratio 1:2, then h=\frac{2x_{1}+x_{2}}{3} and k=\frac{\frac{2}{x_{1}}+\frac{1}{x_{2}}}{3}=\frac{2x_{2}+x_{1}}{3x_{1}x_{2}}, that is, \Longrightarrow 2x_{1}+x_{2}=3h…call this equation I.

and x_{1}+2x_{2}=3(-\frac{1}{4})k=(-\frac{3}{4})k….call this equation II.

Adding I and II, we get 3(x_{1}+x_{2})=3(h-\frac{k}{4}), that is,

3(-\frac{c}{4})=3(h-\frac{k}{4}) \Longrightarrow (h-\frac{k}{4})=-\frac{c}{4}….call this equation III.

Subtracting II from I, we get x_{1}-x_{2}=3(h+\frac{k}{4})

\Longrightarrow (x_{1}-x_{2})^{2}=9(h+\frac{k}{4})^{2}

\Longrightarrow \frac{c^{2}}{16} + 1= 9(h+\frac{k}{4})^{2}

\Longrightarrow (h-\frac{k}{4})^{2}+1=9(h+\frac{k}{4})^{2}

\Longrightarrow h^{2}-\frac{1}{2}hk+\frac{k^{2}}{16}+1=9(h^{2}+\frac{1}{2}hk+\frac{k^{2}}{16})

\Longrightarrow 16h^{2}+10hk+k^{2}-2=0

so that the locus of R(h,k) is 16x^{2}+10xy+y^{2}-2=0

More later,

Nalin Pithwa.

Prioritize your passions and commitments, says Dr. Shawna Pandya

Kids now-a-days need counselling for a choice of career. In my humble opinion, excellence in any field of knowledge/human endeavour gives deep satisfaction as well as a means of livelihood. But, I am a mere mortal, most of my students are bright, ambitious, multi-talented and hard-working. I would like to present to them the views of one of my “idols”, though not a mathematician…

Dr. Shawna Pandya:

Hats off to Dr. Shawna Pandya, belated though from me…:-)

— Nalin Pithwa.

Cartesian system and straight lines: IITJEE Mains: Problem solving skills

Problem 1:

The line joining A(b\cos{\alpha},b\sin{\alpha}) and B(a\cos{\beta},a\sin{\beta}) is produced to the point M(x,y) so that AM:MB=b:a, then find the value of x\cos{\frac{\alpha+\beta}{2}}+y\sin{\frac{\alpha+\beta}{2}}.

Solution 1:

As M divides AB externally in the ratio b:a, we have x=\frac{b(a\cos{\beta})-a(b\cos{\alpha})}{b-a} and y=\frac{b(a\sin{\beta})-a(b\sin{\alpha})}{b-a} which in turn

\Longrightarrow \frac{x}{y} = \frac{\cos{\beta}-cos{\alpha}}{\sin{\beta}-\sin{\alpha}}

= \frac{2\sin{\frac{\alpha+\beta}{2}}\sin{\frac{\alpha-\beta}{2}}}{2\cos{\frac{\alpha+\beta}{2}}\sin{\frac{\beta-\alpha}{2}}}

\Longrightarrow x\cos{\frac{\alpha+\beta}{2}}+y\sin{\frac{\alpha+\beta}{2}}=0

Problem 2:

If the circumcentre of a triangle lies at the origin and the centroid in the middle point of the line joining the points (a^{2}+1,a^{2}+1) and (2a,-2a), then where does the orthocentre lie?

Solution 2:

From plane geometry, we know that the circumcentre, centroid and orthocentre of a triangle lie on a line. So, the orthocentre of the triangle lies on the line joining the circumcentre (0,0) and the centroid (\frac{(a+1)^{2}}{2},\frac{(a-1)^{2}}{2}), that is, y.\frac{(a+1)^{2}}{2} = x.\frac{(a-1)^{2}}{2}, or (a-1)^{2}x-(a+1)^{2}y=0. That is, the orthocentre lies on this line.

Problem 3:

If a, b, c are unequal and different from 1 such that the points (\frac{a^{3}}{a-1},\frac{a^{2}-3}{a-1}), (\frac{b^{3}}{b-1},\frac{b^{2}-3}{b-1}) and (\frac{c^{3}}{c-1},\frac{c^{2}-3}{c-1}) are collinear, then which of the following option is true?

a: bc+ca+ab+abc=0

b: a+b+c=abc

c: bc+ca+ab=abc

d: bc+ca+ab-abc=3(a+b+c)

Solution 3:

Suppose the given points lie on the line lx+my+n=0 then a, b, c are the roots of the equation :

lt^{3}+m(t^{2}-3)+n(t-1)=0, or


\Longrightarrow a+b+c=-\frac{m}{l} and ab+bc+ca=\frac{n}{l}, that is, abc=(3m+n)/l

Eliminating l, m, n, we get abc=-3(a+b+c)+bc+ca+ab

\Longrightarrow bc+ca+ab-abc=3(a+b+c), that is, option (d) is the answer.

Problem 4:

If p, x_{1}, x_{2}, \ldots, x_{i}, \ldots and q, y_{1}, y_{2}, \ldots, y_{i}, \ldots are in A.P., with common difference a and b respectively, then on which line does the centre of mean position of the points A_{i}(x_{i},y_{i}) with i=1,2,3 \ldots, n lie?

Solution 4:

Note: Centre of Mean Position is (\frac{\sum{xi}}{n},\frac{\sum {yi}}{n}).

Let the coordinates of the centre of mean position of the points A_{i}, i=1,2,3, \ldots,n be (x,y) then

x=\frac{x_{1}+x_{2}+x_{3}+\ldots + x_{n}}{n} and y=\frac{y_{1}+y_{2}+\ldots + y_{n}}{n}

\Longrightarrow x = \frac{np+a(1+2+\ldots+n)}{n}, y=\frac{nq+b(1+2+\ldots+n)}{n}

\Longrightarrow x=p+ \frac{n(n+1)}{2n}a and y=q+ \frac{n(n+1)}{2n}b

\Longrightarrow x=p+\frac{n+1}{2}a, and y=q+\frac{n+1}{2}b

\Longrightarrow 2\frac{(x-p)}{a}=2\frac{(y-q)}{b} \Longrightarrow bx-ay=bp-aq, that is, the CM lies on this line.

Problem 5:

The line L has intercepts a and b on the coordinate axes. The coordinate axes are rotated through a fixed angle, keeping the origin fixed. If p and q are the intercepts of the line L on the new axes, then what is the value of \frac{1}{a^{2}} - \frac{1}{p^{2}} + \frac{1}{b^{2}} - \frac{1}{q^{2}}?

Solution 5:

Equation of the line L in the two coordinate systems is \frac{x}{a} + \frac{y}{b}=1, and \frac{X}{p} + \frac{Y}{q}=1 where (X,Y) are the new coordinate of a point (x,y) when the axes are rotated through a fixed angle, keeping the origin fixed. As the length of the perpendicular from the origin has not changed.

\frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}}=\frac{1}{\sqrt{\frac{1}{p^{2}} + \frac{1}{q^{2}}}}

\Longrightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{p^{2}} + \frac{1}{q^{2}}

or \frac{1}{a^{2}} - \frac{1}{p^{2}} + \frac{1}{b^{2}} - \frac{1}{q^{2}}=0. So, the value is zero.

Problem 6:

Let O be the origin, A(1,0) and B(0,1) and P(x,y) are points such that xy>0 and x+y<1, then which of the following options is true:

a: P lies either inside the triangle OAB or in the third quadrant

b: P cannot lie inside the triangle OAB

c: P lies inside the triangle OAB

d: P lies in the first quadrant only.

Solution 6:

Since xy>0, P either lies in the first quadrant or in the third quadrant. The inequality x+y<1 represents all points below the line x+y=1. So that xy>0 and x+y<1 imply that either P lies inside the triangle OAB or in the third quadrant.

Problem 7:

An equation of a line through the point (1,2) whose distance from the point A(3,1) has the greatest value is :

option i: y=2x

option ii: y=x+1

option iii: x+2y=5

option iv: y=3x-1

Solution 7:

Let the equation of the line through (1,2) be y-2=m(x-1). If p denotes the length of the perpendicular from (3,1) on this line, then p=|\frac{2m+1}{\sqrt{m^{2}+1}}|

\Longrightarrow p^{2}=\sqrt{\frac{4m^{2}+4m+1}{m^{2}+1}}=4+ \frac{4m-3}{m^{2}+1}=s, say

then p^{2} is greatest if and only if s is greatest.

Now, \frac{ds}{dm} = \frac{(m^{2}+1)(4)-2m(4m-3)}{(m^{2}+1)^{2}} = \frac{-2(2m-1)(m-2)}{(m^{2}+1)^{2}}

\frac{ds}{dm} = 0 so that \Longrightarrow m = \frac{1}{2}, 2. Also, \frac{ds}{dm}<0, if m<\frac{1}{2}, and

\frac{ds}{dm} >0, if 1/2<m<2

and \frac{ds}{dm} <0, if m>2. So s is greatest for m=2. And, thus, the equation of the required line is y=2x.

Problem 8:

The points A(-4,-1), B(-2,-4), Slatex C(4,0)$ and D(2,3) are the vertices of a :

option a: parallelogram

option b: rectangle

option c: rhombus

option d: square.

Note: more than one option may be right. Please mark all that are right.

Solution 8:

Mid-point of AC = (\frac{-4+4}{2},\frac{-1+0}{2})=(0, \frac{-1}{2})

Mid-point of BD = (\frac{-2+2}{2},\frac{-4+3}{2})=(0,\frac{-1}{2})

\Longrightarrow the diagonals AC and BD bisect each other.

\Longrightarrow ABCD is a parallelogram.

Next, AC= \sqrt{(-4-4)^{2}+(-1+0)^{2}}=\sqrt{64+1}=\sqrt{65} and BD=\sqrt{(-2-2)^{2}+(-4+3)^{2}}=\sqrt{16+49}=\sqrt{65} and since the diagonals are also equal, it is a rectangle.

As AB=\sqrt{(-4+2)^{2}+(-1+4)^{2}}=\sqrt{13} and BC=\sqrt{(-2-4)^{2}+(-4)^{2}}=\sqrt{36+16}=sqrt{52}, the adjacent sides are not equal and hence, it is neither a rhombus nor a square.

Problem 9:

Equations (b-c)x+(c-a)y+(a-b)=0 and (b^{3}-c^{3})x+(c^{3}-a^{3})y+a^{3}-b^{3}=0 will represent the same line if

option i: b=c

option ii: c=a

option iii: a=b

option iv: a+b+c=0

Solution 9:

The two lines will be identical if there exists some real number k, such that

b^{3}-c^{3}=k(b-c), and c^{3}-a^{3}=k(c-a), and a^{3}-b^{3}=k(a-b).

\Longrightarrow b-c=0 or b^{2}+c^{2}+bc=k

\Longrightarrow c-a=0 or c^{2}+a^{2}+ac=k, and

\Longrightarrow a-b=0 or a^{2}+b^{2}+ab=k

That is, b=c or c=a, or a=b.

Next, b^{2}+c^{2}+bc=c^{2}+a^{2}+ca \Longrightarrow b^{2}-a^{2}=c(a-b). Hence, a=b, or a+b+c=0.

Problem 10:

The circumcentre of a triangle with vertices A(a,a\tan{\alpha}), B(b, b\tan{\beta}) and C(c, c\tan{\gamma}) lies at the origin, where 0<\alpha, \beta, \gamma < \frac{\pi}{2} and \alpha + \beta + \gamma = \pi. Show that it’s orthocentre lies on the line 4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}x-4\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}y=y

Solution 10:

As the circumcentre of the triangle is at the origin O, we have OA=OB=OC=r, where r is the radius of the circumcircle.

Hence, OA^{2}=r^{2} \Longrightarrow a^{2}+a^{2}\tan^{2}{\alpha}=r^{2} \Longrightarrow a = r\cos{\alpha}

Therefore, the coordinates of A are (r\cos{\alpha},r\sin{\alpha}). Similarly, the coordinates of B are (r\cos{\beta},r\sin{\beta}) and those of C are (r\cos{\gamma},r\sin{\gamma}). Thus, the coordinates of the centroid G of \triangle ABC are


Now, if P(h,k) is the orthocentre of \triangle ABC, then from geometry, the circumcentre, centroid, and the orthocentre of a triangle lie on a line, and the slope of OG equals the slope of OP.

Hence, \frac{\sin{\alpha}+\sin{\beta}+\sin{\gamma}}{\cos{\alpha}+\cos{\beta}+\cos{\gamma}}=\frac{k}{h}

\Longrightarrow \frac{4\cos{(\frac{\alpha}{2})}\cos{(\frac{\beta}{2})}\cos{(\frac{\gamma}{2})}}{1+4\sin{(\frac{\alpha}{2})}\sin{(\frac{\beta}{2})}\sin{(\frac{\gamma}{2})}}= \frac{k}{h}

because \alpha+\beta+\gamma=\pi.

Hence, the orthocentre P(h,k) lies on the line


Hope this gives an assorted flavour. More stuff later,

Nalin Pithwa.

Frenchmen and mathematics ! :-) :-) :-)

Mathematicians are like Frenchmen: whatever you tell to them they translate in their own language and forthwith it is something entirely different. — GOETHE.

🙂 🙂 🙂