mind blowing questions....@quant

[khanna_sumit]

srihari, friend ur logic is still wrong....... u are assuming that the conditions are same for 2-D and 3-D, which is not the case actually...things are different here...luckily, in previous attempt u hit the correct answer but ur logic was wrong....answer is still 5/27.....but why?????????for a hint if i can find the ratio of radii of the 3 types of sphere then problem is solved.....visualisation will definitely solve this problem........i dont want to rob u people of the pleasure of solving this prob therefore im not posting the solution....best of luck..............

[khanna_sumit]

Originally Posted by khanna_sumit

srihari, friend ur logic is still wrong....... u are assuming that the conditions are same for 2-D and 3-D, which is not the case actually...things are different here...luckily, in previous attempt u hit the correct answer but ur logic was wrong....answer is still 5/27.....but why?????????for a hint if i can find the ratio of radii of the 3 types of sphere then problem is solved.....visualisation will definitely solve this problem........i dont want to rob u people of the pleasure of solving this prob therefore im not posting the solution....best of luck..............

here is another superb analytical reasoning question.....hand picked for phylomaths.......

there are 12 lookalike balls....11 of these balls have some weight but weight of 1 of the balls is different from the other set of 11......we dont know whether it is heavier than the other balls or lighter that the others.......u are presented with a weighing balance and no weights......devise a system of three weighings by which u can identify which one is the odd ball and whether it is heavier or lighter than the other balls.......

[khanna_sumit]

Originally Posted by assignus

how about this...

Q) If a point on the cimcumsphere is chosen randomly, then what is the probability that it lies on on the five inspheres ??

any answers??

yes assignus...... answer in that case will be=surface area of the five inner spheres/volume of the circumsphere......for that also u will have to find the ratio of radiii only..........

[assignus]

Originally Posted by khanna_sumit

yes assignus...... answer in that case will be=surface area of the five inner spheres/volume of the circumsphere......for that also u will have to find the ratio of radiii only..........

Nop....answer in this case would be

[Surface Area common to 5 inspheres and the circumsphere (actually only 4 cuz inner most one never touches the cicumsphere)] divided by [surface area of the circumsphere]

by how do we calculate the area of the insphere which is touching the circumsphere? i believe it will be only a point at which both the surfaces touch.....so totally [4 points] (for 4 inspheres) divided by [no. of points of circum sphere ] ...can anyone count that for me???

any inputs ppl??

[assignus]

Originally Posted by khanna_sumit

here is another superb analytical reasoning question.....hand picked for phylomaths.......

there are 12 lookalike balls....11 of these balls have some weight but weight of 1 of the balls is different from the other set of 11......we dont know whether it is heavier than the other balls or lighter that the others.......u are presented with a weighing balance and no weights......devise a system of three weighings by which u can identify which one is the odd ball and whether it is heavier or lighter than the other balls.......

4 steps.....couldnt figure out in 3 steps

divide them into 3 groups of 4 each.... OOOO .. OOOO .. OOOO
weight any two

if they are equal ............then the other 4 has the faulty ball.... OOOO ....pick any 2 from them ...if they are equal (2nd weighing) then the remaining 2 has the faulty ball ..OO....pick a ball from this and weigh (3rd weighing) with the good ball....if it is equal then the other one the faulty ball.....havd to do 4th weighing to find out whether it is lighter or heavier?

if they are not equal.........take the lighter 4 ball group and weigh (2nd weighing) with the other 4 balls...... if they are equal then the first 4 has the faulty ball which is heavier...... if they are not equal then the lighter group has the faulty ball which is lighter....... now we came down to 4 balls and we also know whether the faulty ball is lighter or heavier....

case 1: 4 balls and we know it has a lighter faulty ball......OOOO ..divide into 2 groups OO .. OO weigh them (3rd weighing) ......find out which is lighter ....then pick one from that and weigh (4th weighing) it with a good ball.....if it is lighter then that the ball...other wise its the other one...

case 2: 4 balls and we know it has a heavier ball....same procedure as in case 1...

pls tell how to do in 3 steps...

[khanna_sumit]

Originally Posted by assignus

Nop....answer in this case would be

[Surface Area common to 5 inspheres and the circumsphere (actually only 4 cuz inner most one never touches the cicumsphere)] divided by [surface area of the circumsphere]

by how do we calculate the area of the insphere which is touching the circumsphere? i believe it will be only a point at which both the surfaces touch.....so totally [4 points] (for 4 inspheres) divided by [no. of points of circum sphere ] ...can anyone count that for me???

any inputs ppl??

bravo eccentric....correct answer......since at least one has answered it so i'll post my solution for it now........

from the symmetry of the problem we can say that the centre of both insphere and circumsphere will be same and that center will be equidistant from the four vertices and four faces as well........also the center will be the point of intersection of the four altitudes of the tetrahedron..means it lies on all the altitudes.....friends, lets visualise from here....i'll directly tell u what to visualise.... visualise a tetrahedron with one of the altitudes drawn...this altitude joins the centroid of the base with the opposite vertex.......now start visualising that this centroid of the base is rising gradually on the altitude to reach at the center of circumsphere which lies somewhere on the altitude.....u will be able to see some planes rising with this centroid... this will further help u to visualise four smaller tetrahedra being formed each of which will have the center of the circumsphere as its vertex........and these four tetrahedra will be congruent, each having volume 1/4 th of the volume of the original tetrahedron.....volume depends upon product of base and altitude.......comparing the volumes of original and smaller tetrahedra, we see that the bases of both are same therefore since volume ratio is 1:4, ratio of heights should also be 1:4......so height of smaller tet.. is 1/4th of the altitude of the original tet.......therefore the center divides the altitude in ratio 1:3, which form circum radius and inradius.....so ration of circumradius and inradius is 3:1.......using this only i can prove that the diameter of the third type of sphere will be 3-1=2 hence radius of that will be 1....so radii ratio is 1:1:3....hence volume ratio is 1:1:27......therefore 5*1/27 is the required probability.....soon i'll be posting elaborated solution with details....... and diagram as well.....and for ur question assignus.....since radii ratio is 1:1:3 therefore surface srea of a smaller sphere is 4/3(pi*(1)^2).......

[khanna_sumit]

Originally Posted by Eccentric

Referring to image:


Or alternatively you can view image at:
http://img182.echo.cx/img182/5145/trtrahedron8pt.jpg


In here I'll try to derive the circu-radius and In-radius for Tetrahedron with following premises (this I don't think we need to prove):
1. The altitude form the top vertex intersect the bottom face (Equilateral Triangle) at its centroid.
2. The circum-center lies on the altitude defined above.


Let side of a face of TetraHedron ABCD = 1
BL (altitude to Triangle BCD) = sqrt(3)/2
BO = circumradius of Triangle BCD = 2/3 * BL = 1/sqrt(3)

In right triangle ABO (right angled at O)
AB = 1
AO^2 = AB^2-BO^2
AO = sqrt(2/3)

Now we need to find point P on line AO such that AP = BP (then P will be the cicumcenter of the tetrahedron)
let OP = x
AP^2 = (AO- BO)^2
BP^2 = OP^2 + BO^2
and AP^2 = BP^2
(sqrt(2/3) - x)^2 = x^2 + 1/3
2/3 - 2*x* sqrt(2/3) = 1/3
x= 1/6*sqrt(3/2) = 1/6*sqrt(6/4)=sqrt(6)/12

AP = AO - x
AO = sqrt(2/3) = sqrt(6/9)=sqrt(6)/3
AP = sqrt(6)/4

hence cicumradius = sqrt(6)/4
OP will become inradius as this is perpandicular distance of a face from centerof tetrahedron, radius(incirlce) = sqrt(6)/12


Small sphere diameter = circumradius-inradius (this will happen when you drop a perpandicular from center to either face) = sqrt(6)/6
radius (small circale) = sqrt(6)/12



probability = (4* radius (small circale)^3 + radius(incircle)^3)/ circumradius ^3 = 5/27
(Taking volumes)

bravo eccentric....correct answer......since at least one has answered it so i'll post my solution for it now........

from the symmetry of the problem we can say that the centre of both insphere and circumsphere will be same and that center will be equidistant from the four vertices and four faces as well........also the center will be the point of intersection of the four altitudes of the tetrahedron..means it lies on all the altitudes.....friends, lets visualise from here....i'll directly tell u what to visualise.... visualise a tetrahedron with one of the altitudes drawn...this altitude joins the centroid of the base with the opposite vertex.......now start visualising that this centroid of the base is rising gradually on the altitude to reach at the center of circumsphere which lies somewhere on the altitude.....u will be able to see some planes rising with this centroid... this will further help u to visualise four smaller tetrahedra being formed each of which will have the center of the circumsphere as its vertex........and these four tetrahedra will be congruent, each having volume 1/4 th of the volume of the original tetrahedron.....volume depends upon product of base and altitude.......comparing the volumes of original and smaller tetrahedra, we see that the bases of both are same therefore since volume ratio is 1:4, ratio of heights should also be 1:4......so height of smaller tet.. is 1/4th of the altitude of the original tet.......therefore the center divides the altitude in ratio 1:3, which form circum radius and inradius.....so ration of circumradius and inradius is 3:1.......using this only i can prove that the diameter of the third type of sphere will be 3-1=2 hence radius of that will be 1....so radii ratio is 1:1:3....hence volume ratio is 1:1:27......therefore 5*1/27 is the required probability.....soon i'll be posting elaborated solution with details.......

[Aarav]

Seem some geometry studs out there.

OK solve...

A point P inside the equilateral triangle is of length 3,4,5 from the vertices. Find the area of the triangle.

3 medians are of length 3,4,5. Find the area of the triangle.

[Eccentric]

Originally Posted by khanna_sumit

here is another superb analytical reasoning question.....hand picked for phylomaths.......

there are 12 lookalike balls....11 of these balls have some weight but weight of 1 of the balls is different from the other set of 11......we dont know whether it is heavier than the other balls or lighter that the others.......u are presented with a weighing balance and no weights......devise a system of three weighings by which u can identify which one is the odd ball and whether it is heavier or lighter than the other balls.......

Sumit,
Please post new problem on a new thread. I'll post the answer for three weighings.

[Eccentric]

Originally Posted by assignus

how about this...

Q) If a point on the cimcumsphere is chosen randomly, then what is the probability that it lies on on the five inspheres ??

any answers??

zero ! ! ! !