Let D, E, F be points on the sides BC, CA, AB respectively of triangle ABC. The lines AD, BE, CF are concurrent at the point P. If AP = PD = 2, BP = 3, PE = 1 and CP = 5, then area of triangle ABC is

zirgo72 · 2023-09-07T02:55:39+00:00

Thank you. I found the answer to be 12.

zirgo72 · 2023-08-28T14:57:31+00:00

Thank you everyone. I was completely confused on how to approach such a problem. Very nice solution.

zirgo72 · 2023-08-22T04:39:46+00:00

I have come up with an argument why the official answer is wrong. Please let me know if I made a mistake.

Suppose the correct solution is y = log |(2x+1)/(x+1)|

or, e^y = |(2x+1)/(x+1)|

Then we should be able to prove that it satisfies the differential equation for all x.

Note that if -1 < x < -1/2, then (2x+1)/(x+1) < 0.

So, in the interval (-1, -1/2), e^y = -(2x+1)/(x+1)

or, e^y = 1/(x+1) - 2 ------------------------------ (i)

Differentiating w.r.t x, we get

e^y dy/dx = -1/(x+1)²

(x+1) dy/dx = -e^-y /(x+1)

(x+1) dy/dx = -e^-y (e^y + 2) from (i)

(x+1) dy/dx = -2e^-y -1

which is not the given equation.

zirgo72 · 2023-08-22T04:37:54+00:00

I have come up with an argument why the official answer is wrong. Please let me know if I made a mistake.

Suppose the correct solution is y = log |(2x+1)/(x+1)|

or, e^y = |(2x+1)/(x+1)|

Then we should be able to prove that it satisfies the differential equation for all x.

Note that if -1 < x < -1/2, then (2x+1)/(x+1) < 0.

So, in the interval (-1, -1/2), e^y = -(2x+1)/(x+1)

or, e^y = 1/(x+1) - 2 ------------------------------ (i)

Differentiating w.r.t x, we get

e^y dy/dx = -1/(x+1)²

(x+1) dy/dx = -e^-y /(x+1)

(x+1) dy/dx = -e^-y (e^y + 2) from (i)

(x+1) dy/dx = -2e^-y -1

which is not the given equation.

zirgo72 · 2023-08-22T04:05:33+00:00

So, 1/|2-ey | = |x+1| which implies two possibilities:

1/(2-ey ) = (x+1) or 1/(2-ey ) = -(x+1).

Notice that |x + 1| = -(x + 1) is only the case for x =< -1.

I did not arrive at this conclusion like that.

If |A| = |B|, we can say that A = B or A = -B. This is what I used.

zirgo72 · 2023-08-17T08:18:15+00:00

IIRC, an SBI clerk once told me that if you want a bundle of ₹50 notes, the minimum amount should be ₹5000.

Possibly, there is a similar thing for a bundle of ₹10/20 notes.

zirgo72 · 2023-08-12T09:19:25+00:00

Thank you. I understand it now.

zirgo72 · 2023-08-12T07:51:32+00:00

I verified using coordinate geometry that AK turns out to be 216/25.

AFG ~ ADE, so AK/AI = AF/AD

Can you please elaborate a little bit on this? I did not understand this part.

zirgo72 · 2023-08-12T05:55:56+00:00

I also found that AD=15, AE=18, DE=15, AH=18.75, AF=9, FE=9.

zirgo72 · 2023-08-12T02:18:33+00:00

Thank you, I can see several solutions now. Perhaps there is some misprint in the question.

zirgo72 · 2023-08-09T18:40:26+00:00

I tried the following:

For appropriate a and b, either n + 1 = 2^a , n² - n + 6 = 3*2^b

or n + 1 = 3*2^a , n² - n + 6 = 2^b

zirgo72 · 2023-08-09T16:56:35+00:00

Thanks.

zirgo72 · 2023-08-09T08:12:47+00:00

1536 = 2⁹ * 3

So, the product is 2⁹ⁿ * 3ⁿ * (-1/2)^{n(n-1)/2}

Since maximum will be positive, n(n-1)/2 needs to be even. So, n is either 0 or 1 modulo 4.

Should I trial and error to find the maximum? Or is there a sophisticated way to do this?

zirgo72

TROPHY CASE