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Salman Khan solves IIT JEE 2010 problems
by sambarvada Sun May 29, 2011 11:59 pm
http://www.khanacademy.org/video/iit-jee-trigonometry-problem-1?playlist=IIT%20JEE%20Questions
sambarvada- Posts : 585
Join date : 2011-04-29
Re: Salman Khan solves IIT JEE 2010 problems
by Guest Mon May 30, 2011 1:07 am
thanks. i was able to see a JEE problem up close and study how difficult/easy it is.
but his solution will land the student in trouble. the shortest solution is this (as a reader comments):
but his solution will land the student in trouble. the shortest solution is this (as a reader comments):
Suggestion to do this problem more quickly...
take the three angles as
A=a B=a+d C=a-d since A+B+C=180 u get a=60 d=30 being a 30-60-90 triangle you can apply its propeties ie their side ratios
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Re: Salman Khan solves IIT JEE 2010 problems
by sambarvada Mon May 30, 2011 8:52 am
Huzefa Kapasi wrote:thanks. i was able to see a JEE problem up close and study how difficult/easy it is.
but his solution will land the student in trouble. the shortest solution is this (as a reader comments):Suggestion to do this problem more quickly...
take the three angles as
A=a B=a+d C=a-d since A+B+C=180 u get a=60 d=30 being a 30-60-90 triangle you can apply its propeties ie their side ratios
===> Did you mean to say that C=a+2d?
sambarvada- Posts : 585
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Re: Salman Khan solves IIT JEE 2010 problems
by sambarvada Mon May 30, 2011 9:16 am
Huzefa Kapasi wrote:thanks. i was able to see a JEE problem up close and study how difficult/easy it is.
but his solution will land the student in trouble. the shortest solution is this (as a reader comments):Suggestion to do this problem more quickly...
take the three angles as
A=a B=a+d C=a-d since A+B+C=180 u get a=60 d=30 being a 30-60-90 triangle you can apply its propeties ie their side ratios
===> B=60
But how are you so sure that A=30 and B=90?
sambarvada- Posts : 585
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Re: Salman Khan solves IIT JEE 2010 problems
by Guest Mon May 30, 2011 9:23 am
sambarvada wrote:
===> Did you mean to say that C=a+2d?
i am not sure how you got this. C is an angle. a and d are lengths of the sides.
what i (the reader) meant was that if we assume that it is a right angled triangle and that the hypotenuse has a length of 1, then all the sides and angles are known immediately (hypotenuse = 1, base = 1/2, perpendicular = (sqr. rt. 3 )/ 2). so the sines and cosines be easily calculated of the angles in the sum (or you can convert the sines and cosines into ratio of sides; sine theta = p/h etc.).
*some people have curly brown hair turned permanently black*
*sine p h cos b h tan p b*
Last edited by Huzefa Kapasi on Mon May 30, 2011 9:30 am; edited 3 times in total
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Re: Salman Khan solves IIT JEE 2010 problems
by Guest Mon May 30, 2011 9:28 am
sambarvada wrote:But how are you so sure that A=30 and B=90?
it is the similarity that matters. the function given in the question will compute to the same for all similar triangles. by "similar" i mean ratios given in the question (and not similar triangles). evidently the function is not for any one particular triangle as evident in the salman khan solution -- there can be many such triangles. but the function will always compute to the same.
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Re: Salman Khan solves IIT JEE 2010 problems
by Guest Mon May 30, 2011 9:33 am
in fact that reader's solution is very smart. though not elegant, it shows how the salman khan solution is a total waste of time (read IIT admission denied).
Guest- Guest
To H
by sambarvada Mon May 30, 2011 9:36 am
Look at your first reply.
You said Suggestion to do this problem more quickly...
take the three angles as
A=a B=a+d C=a-d since A+B+C=180 u get a=60 d=30 being a 30-60-90 triangle you can apply its propeties ie their side ratios.
Tell me what you meant by A=a B=a+d C=a-d?
if a is supposed to denote the length of the side opposite to angle A, then how can angle A=a?
You said Suggestion to do this problem more quickly...
take the three angles as
A=a B=a+d C=a-d since A+B+C=180 u get a=60 d=30 being a 30-60-90 triangle you can apply its propeties ie their side ratios.
Tell me what you meant by A=a B=a+d C=a-d?
if a is supposed to denote the length of the side opposite to angle A, then how can angle A=a?
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Re: Salman Khan solves IIT JEE 2010 problems
by Guest Mon May 30, 2011 9:50 am
1) it is not my reply. it is the reply by a reader in the salman khan video;
2) the reader typed in a hurry but the meaning is clear.
what the reader meant was this:
take the three angles as
A=A B=A+N C=A-N since A+B+C=180 u get A=60 N=30 being a 30-60-90 triangle...
in the above he does not get "N=30" but he ASSUMES this for a quicker solution.
2) the reader typed in a hurry but the meaning is clear.
what the reader meant was this:
take the three angles as
A=A B=A+N C=A-N since A+B+C=180 u get A=60 N=30 being a 30-60-90 triangle...
in the above he does not get "N=30" but he ASSUMES this for a quicker solution.
Guest- Guest
Re: Salman Khan solves IIT JEE 2010 problems
by MaxEntropy_Man Mon May 30, 2011 10:12 am
i haven't see khan's solution yet. here's how i reasoned it out. there may be a more elegant solution.
the angles are in arithmetic progression.
so the angles are A, B(=A+d), C=(A+2d).
we know A+B+C = 180 deg. thus, A+A+d+A+2d =3(A+d)=180 deg.
=> A+d = 60 deg.
A and A+d are equidistant in angular measure from 60, but their values should also satisfy A+B+C = 180 deg. the only values that work are A=30 deg and C = 90 deg. from there on it is straightforward.
(a/c)sin[2C]+(c/a)sin[2A] = (a/c)sin[180]+(c/a)sin[60]=0+(c/a)(Sqrt[3]/2).
once we know the angles are 30, 60, and 90 deg, we know that c/a=csc[A]=1/sin[A]=2.
thus the required quantity is 2(Sqrt[3]/2)=Sqrt[3].
eta: this was a recent JEE problem?
if so, !
confirms my suspicion that it has gotten easier over the years.
the angles are in arithmetic progression.
so the angles are A, B(=A+d), C=(A+2d).
we know A+B+C = 180 deg. thus, A+A+d+A+2d =3(A+d)=180 deg.
=> A+d = 60 deg.
A and A+d are equidistant in angular measure from 60, but their values should also satisfy A+B+C = 180 deg. the only values that work are A=30 deg and C = 90 deg. from there on it is straightforward.
(a/c)sin[2C]+(c/a)sin[2A] = (a/c)sin[180]+(c/a)sin[60]=0+(c/a)(Sqrt[3]/2).
once we know the angles are 30, 60, and 90 deg, we know that c/a=csc[A]=1/sin[A]=2.
thus the required quantity is 2(Sqrt[3]/2)=Sqrt[3].
eta: this was a recent JEE problem?
if so, !
confirms my suspicion that it has gotten easier over the years.
Last edited by MaxEntropy_Man on Mon May 30, 2011 10:18 am; edited 1 time in total
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Re: Salman Khan solves IIT JEE 2010 problems
by Guest Mon May 30, 2011 10:16 am
MaxEntropy_Man wrote:
A and A+d are equidistant in angular measure from 60, but their values should also satisfy A+B+C = 180 deg. the only values that work are A=30 deg and C = 90 deg. from there on it is straightforward.
what about 29, 60, 91? or 25, 60, 95?
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Re: Salman Khan solves IIT JEE 2010 problems
by MaxEntropy_Man Mon May 30, 2011 10:19 am
i was wrong about the only solution that works is the one i came up with. but if the quantity is unique, it should be possible to prove it more generally than for a specific set of angular measures. let me think some more.
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Re: Salman Khan solves IIT JEE 2010 problems
by Guest Mon May 30, 2011 10:25 am
MaxEntropy_Man wrote:eta: this was a recent JEE problem?
if so, !
confirms my suspicion that it has gotten easier over the years.
*ahem* it's nice to begin the day with a pat on one's back.
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Re: Salman Khan solves IIT JEE 2010 problems
by MaxEntropy_Man Mon May 30, 2011 12:19 pm
more general solution:
my proof that B=A+d = 60 deg still stands. i start the rest of the proof here.
the quantity to be evaluated is
(a/c)sin[2C]+(c/a)sin[2A] ---- Eq (1)
now using the sine rule we know that for any triangle,
a/sin[A] = b/sin[B] = c/sin[C] ----- Eq(2)
using this,
a/c = sin[A]/sin[C], c/A = sin[C]/sin[A] ----Eq (3a, 3b)
substituting equations 3a and 3b in equation 1 and expanding sin[2C] as
2 sin[C]cos[C] and sin[2A] similarly, the given quantity becomes
2(sin[A]cos[C]+cos[A]sin[C])
but this is nothing but 2sin[A+C] (using the sine of sum of angles formula); but A+C = 120 deg, since B = 60 deg.
thus the given quantity is 2sin[A+C] = 2 sin[120] = 2 Sqrt[3]/2 = Sqrt[3]
if this was a multiple choice q, i'd still use my original guess. but if the question asked for a more general proof, then the method i give in this post works.
eta: i still stand by my original statement that a q like this would have no place in a JEE from a prior era. too simple.
my proof that B=A+d = 60 deg still stands. i start the rest of the proof here.
the quantity to be evaluated is
(a/c)sin[2C]+(c/a)sin[2A] ---- Eq (1)
now using the sine rule we know that for any triangle,
a/sin[A] = b/sin[B] = c/sin[C] ----- Eq(2)
using this,
a/c = sin[A]/sin[C], c/A = sin[C]/sin[A] ----Eq (3a, 3b)
substituting equations 3a and 3b in equation 1 and expanding sin[2C] as
2 sin[C]cos[C] and sin[2A] similarly, the given quantity becomes
2(sin[A]cos[C]+cos[A]sin[C])
but this is nothing but 2sin[A+C] (using the sine of sum of angles formula); but A+C = 120 deg, since B = 60 deg.
thus the given quantity is 2sin[A+C] = 2 sin[120] = 2 Sqrt[3]/2 = Sqrt[3]
if this was a multiple choice q, i'd still use my original guess. but if the question asked for a more general proof, then the method i give in this post works.
eta: i still stand by my original statement that a q like this would have no place in a JEE from a prior era. too simple.
MaxEntropy_Man- Posts : 14702
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Re: Salman Khan solves IIT JEE 2010 problems
by Marathadi-Saamiyaar Mon May 30, 2011 12:34 pm
This is still too high a standard.
I will have to postpone taking this JEE for a few more years. By that time, the question papers will have 15 wrong/ambiguous questions and the examinees will get full points for those questions.
And, I will then automatically make the cut.
I will have to postpone taking this JEE for a few more years. By that time, the question papers will have 15 wrong/ambiguous questions and the examinees will get full points for those questions.
And, I will then automatically make the cut.
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Re: Salman Khan solves IIT JEE 2010 problems
by Guest Mon May 30, 2011 12:53 pm
MaxEntropy_Man wrote:more general solution:
my proof that B=A+d = 60 deg still stands. i start the rest of the proof here.
the quantity to be evaluated is
(a/c)sin[2C]+(c/a)sin[2A] ---- Eq (1)
now using the sine rule we know that for any triangle,
a/sin[A] = b/sin[B] = c/sin[C] ----- Eq(2)
using this,
a/c = sin[A]/sin[C], c/A = sin[C]/sin[A] ----Eq (3a, 3b)
substituting equations 3a and 3b in equation 1 and expanding sin[2C] as
2 sin[C]cos[C] and sin[2A] similarly, the given quantity becomes
2(sin[A]cos[C]+cos[A]sin[C])
but this is nothing but 2sin[A+C] (using the sine of sum of angles formula); but A+C = 120 deg, since B = 60 deg.
thus the given quantity is 2sin[A+C] = 2 sin[120] = 2 Sqrt[3]/2 = Sqrt[3]
if this was a multiple choice q, i'd still use my original guess. but if the question asked for a more general proof, then the method i give in this post works.
eta: i still stand by my original statement that a q like this would have no place in a JEE from a prior era. too simple.
very nice Il Professore. i think this is shorter than salman khan's solution (but i think salman khan was being overtly simplistic in the video to make even an average student understand). regardless, a commendable effort on you part. or, er, i'll take that back considering that this was an easy sum (though salman khan marvels at the ingenuity of the problem in the video).
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Re: Salman Khan solves IIT JEE 2010 problems
by Guest Mon May 30, 2011 12:58 pm
Huzefa Kapasi wrote:1) it is not my reply. it is the reply by a reader in the salman khan video;
2) the reader typed in a hurry but the meaning is clear.
what the reader meant was this:
take the three angles as
A=A B=A+N C=A-N since A+B+C=180 u get A=60 N=30 being a 30-60-90 triangle...
in the above he does not get "N=30" but he ASSUMES this for a quicker solution.
oops, i just tried to work this problem using the above principles and i ran into an error. if A is assumed to be 60 then B and C become 90 and 120 respectively. it is OK to assume A as 60 but then you have to substitute B and C for the clockwise equivalent in the equation. or, better still, start with assuming A=30 (obviously that is what the reader meant) and then you don't have to substitute the angles in the function. .
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Re: Salman Khan solves IIT JEE 2010 problems
by MaxEntropy_Man Mon May 30, 2011 8:35 pm
Huzefa Kapasi wrote:
i think this is shorter than salman khan's solution.......
yes it is. i use only the sine rule and khan uses the sine and cosine rules.
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