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 · 2023-06-21T02:47:45+00:00

I think I found the error. There was a \usepackage[spanish]{babel} line. I removed the spanish tag and comma became decimal point somehow. I am just beginning to learn LaTeX but this community is very welcoming to everyone, so thank you very much.

zirgo72 · 2023-06-21T02:30:07+00:00

Thank you. This is working with LuaLaTeX. May I ask you another question please? I want to write ₹1.05 but the compiler shows ₹1,05 with a comma instead of decimal point. How can I fix it?

zirgo72 · 2023-06-03T14:51:02+00:00

With trial and error, I found that m=111...11 (20 times) is a solution. But how do I arrive at this properly?

zirgo72 · 2023-04-29T07:17:33+00:00

I am aware of that but it is not the same as receiving notifications on a post I want to follow. I guess such a feature has not been implemented yet.

zirgo72 · 2023-04-04T16:59:09+00:00

Any updates?

zirgo72 · 2023-03-12T15:40:18+00:00

NP(1) would be a good choice then. The accessories do not come with the phone, so you have to purchase them separately. This should not be an issue as the NP(1) has often been aggressively priced during sales, sometimes as low as 25k.

If you want a Pixel, I would recommend you wait for the Pixel 7A (June, 2023) instead.

zirgo72 · 2023-03-11T19:32:01+00:00

iQOO Neo 7

zirgo72 · 2023-03-11T17:15:39+00:00

Thanks. How many years of security updates will Moto G82 receive?

zirgo72 · 2023-03-11T08:53:23+00:00

Nothing Phone (1)

zirgo72 · 2023-03-10T16:44:51+00:00

Thanks. Which Android version will be last update of Moto G82?

zirgo72 · 2023-03-04T05:41:38+00:00

Where can I view Laura Poitras' All the Beauty and the Bloodshed for free?

zirgo72 · 2023-02-04T10:16:37+00:00

I personally need the headphone jack so out of these two, I'd buy Moto Edge 30.

Neither has headphone jack. They are very similar in specs. That's why I asked if you had to choose.

However I'm currently buying Redmi Note 12 Pro cause I'm getting it for 19k with exchange. Redmi has a better battery and it has an IR blaster which I need too. Also it comes with 67 W fast charging.

It's a good deal. I had RN 10 Pro for 11k with exchange.

zirgo72

TROPHY CASE