Salman Khan solves IIT JEE 2010 problems

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

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

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

sambarvada wrote:
===> Did you mean to say that C=a+2d?

sambarvada wrote:But how are you so sure that A=30 and B=90?

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.

MaxEntropy_Man wrote:eta: this was a recent JEE problem?
if so, !
confirms my suspicion that it has gotten easier over the years.

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.

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.

Huzefa Kapasi wrote:
i think this is shorter than salman khan's solution.......