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Intuitively, Yes. A proper proof would look good.

I'm not sure I understood properly. If the line AB is the longest line possible, then won't A and b themselves be the vertices of the triangle?

25 Aug 2014 08:57
in discussion Discussion Forum / Maths » kids in a circle

1 possible arrangement

Re: kids in a circle by , 25 Aug 2014 08:57
25 Aug 2014 08:28
in discussion Discussion Forum / Maths » kids in a circle

i was not ending with GG. I had put dots to represent the pattern is repeating

Re: kids in a circle by , 25 Aug 2014 08:28
25 Aug 2014 04:50
in discussion Discussion Forum / Maths » kids in a circle

GGBBGGBBGG…………..

You cannot start with G and end with G, you have to end with B.
With your reasoning, 1 is true, since divisibility by 4 implies divisibility by 2.

A different question:

If all boys were identical and all girls were identical, how many circular arrangements do we have with the above mentioned condition?

Re: kids in a circle by , 25 Aug 2014 04:50
25 Aug 2014 04:43
in discussion Discussion Forum / Maths » Area between three equal circles

The idea is right. Hope someone completes the details.

Re: Area between three equal circles by , 25 Aug 2014 04:43

If the longest possible line AB is from one boundary (side) to the other, then this will, by default, be the longest side of the triangle.

Clarification required here. You have argued that A and B have to be one of the sides or different sides(you called it 'boundary') but not necessarily the vertex of the triangle. How do we prove that such a line segment is not greater than the longest side of the triangle?

23 Aug 2014 13:19
in discussion Discussion Forum / Maths » Area between three equal circles

First, I will connect the centers of all the 3 circles, which gives an equilateral triangle whose side is of unit 2. I find the area of the triangle and call it Area1. The area of sector in each circle can be calculated (since we know the radius and angle which is 60 degrees) and call it Area2. Now Area1 minus 3 times Area2 given the region between the 3 circles.

Re: Area between three equal circles by , 23 Aug 2014 13:19
23 Aug 2014 12:29
in discussion Discussion Forum / Maths » kids in a circle

The only pattern that satisfies the conditions in the question is GGBBGGBBGG…………..

1. Even number of kids is not enough. For eg: GGBBGG has 6 kids (even) but when they stand in circle 4G's will be together which breaks the rule
2. Yes. The condition holds good only with repeating patterns of GGBB which means multiple of 4
3. Not enough as per the reasoning in point 2.
4. Yes. As per the reasoning in point 2.

Re: kids in a circle by , 23 Aug 2014 12:29

Let the points be A and B.
Keeping B stationary, the longest line AB is possible when A is on the boundary because as long as it is in the interior of the triangle, it can always be moved towards the boundary (or exterior) to make AB longer. If A is kept stationary, the same conclusion can be drawn for B. So both A and B have to be on the boundary for the longest possible line AB.
If the longest possible line AB is from one boundary (side) to the other, then this will, by default, be the longest side of the triangle.

24 Jul 2014 12:37
in discussion Discussion Forum / Chemistry » Redox Reaction

Thanks a lot.. Very well explained

by , 24 Jul 2014 12:37
Kiran (guest) 24 Jul 2014 09:06
in discussion Discussion Forum / Chemistry » Redox Reaction

Good question! The reaction written as above is always a net reaction. In the net reaction its easy to see that Cu2+ has got reduced to Cu metal. Indicating it picked up two electrons. Hydrogen started from oxidation state 0 and ends as +1 - losing one electron per atom of hydrogen. So the net reaction tells you who is the terminal reducing agent and oxidising agent. However the real source of electrons that have come to Cu is hard to find/explain.
Such a description for source of electrons has to come from a detailed and well proved mechanism. Otherwise its impossible to trace the movement of electrons. In the reaction presented above which proceeds without the 'need for a catalyst' it may be a good guess that Hydrogen itself donates electrons and does not involve Oxygen for electron transfer.
However, we can disprove that O2- donates electrons, by realising that water also can act as a source of O2-, and a mixture of, say, Copper sulphate, water and hydrogen does not give rise to Copper metal! That is - no redox reaction occurs.

However if a catalyst is involved, it is safe to say, that the catalyst donates electrons to Cu2+ and then the oxidised catalyst is reduced by hydrogen - thereby oxidising Hydrogen(acts as a reducing agents- means it gets oxidised) and restoring the catalyst to its original state. A 'catalyst' is a catalyst because it gets regenerated and jumps back into the reaction cycle ! Net reaction doesnt show involvement of catalyst, doesnt show consumption and regeneration of catalysts. But only shows terminal oxidants and reduction. These sort of examples are in plenty in the realm of organic chemistry and not so much in inorganic chemistry. Ofcourse its attributed to the fact that it is easier to study organic reaction mechanisms compared to inorganic reaction mechanisms. Which in turn is due to the fact that organic reactions occur at reasonable temperatures and rates of reaction are slower than their inorganic counterparts.

by Kiran (guest), 24 Jul 2014 09:06
22 Jul 2014 08:17
in discussion Discussion Forum / Chemistry » Redox Reaction

CuO + H2 —-> Cu + H2O

In this redox reaction, Cu is gettng reduced and hydrogen is getting oxidized. Does it necessarily imply the electrons which copper is gaining is coming form hydrogen? Can O2- give the electrons to Cu and then form bond with hydrogen? Essentially where are the electrons which copper is gaining is coming from

Redox Reaction by , 22 Jul 2014 08:17
08 Jul 2014 18:40
in discussion Discussion Forum / Maths » kids in a circle

Some kids stand in the form of a circle. For every kid, both his(or her) neighbors are not of same gender. That is, one neighbor of every kid is a boy and the other is a girl. Which of the following are correct (multiple answers could be correct!):

1. There must be even number of kids.
2. The number of kids is divisible by 4.
3. Either there might be even number of boys or even number of girls, but both need not be even.
4. Both, the number of boys and the number of girls is even.

kids in a circle by , 08 Jul 2014 18:40

Prove that distance between any two points of a triangle(a point may be on the boundary or in the interior of the traingle) is not more than the length of the largest side.

Caveat: It might look obvious. Please construct a precise argument.

08 Jul 2014 18:23
in discussion Discussion Forum / Maths » Area between three equal circles

Let $S_1,S_2,S_3$ be three circles each of radius $1$ that touch each other externally. What is the measure of area in between them?

Area between three equal circles by , 08 Jul 2014 18:23
04 Jul 2014 13:29
in discussion Discussion Forum / Maths » 1 = -1

Yes it does. Thank you

Re: 1 = -1 by , 04 Jul 2014 13:29

Consider the set of numbers $\{1,2,3,\dots 2n\}$ for some natural number $n>1$. Among any $n+1$ numbers selected from the set,
prove that

• we can find a pair of numbers so that they do not have common factor (we call them co-prime)
• we can find two numbers $a$ and $b$ such that $a$ divides $b$.
04 Jul 2014 11:29
in discussion Discussion Forum / Maths » A simple question

Correct, thanks for your answer. I would suggest a small correction. You said,

Lets say each of the 5 kids picked a different number of mangoes. This is possible only if they picked : 5, 4, 3, 2, 1 mangoes.

Its better to say:

Suppose each kid picked different number of mangoes. Then, the minimum number of mangoes picked up by them must be $1+2+3+4+5=15$ mangoes.

Re: A simple question by , 04 Jul 2014 11:29
04 Jul 2014 11:27
in discussion Discussion Forum / Maths » 1 = -1

$\sqrt{a}$ is well-defined only if $a\ge 0$. By well-defined, I mean that the function $\sqrt{\cdot}$ takes us from non-negative reals to non-negative reals. Under well-defined conditions, we can prove that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ if and only if $\sqrt{a}$ and $\sqrt{b}$ are well defined. This does not hold for complex numbers as you have illustrated. The flaw is with the assuming that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ holds when at least one of $a$ or $b$ is complex.