Percentages and Triangles

vvr1590 · Senior Member

Percentages and Triangles

webmaster · Administrator

Vinay,

Hey, thanks for getting a screen name. I am extremely busy today but I will get to these questions on my lunch break or later on tonight for sure.

Webmaster

P.S. In the meantime, be sure to look up the section under Miscellaneous the section about right triangles and also look up percentages in the same section.

Date: Jan 18, 2005

i am having trouble on the 1st and last problem you mentioned too...it's not hard..but it takes time...i have always skipped them...for the 2nd and 3rd problem, you can find the trick under the trig section on the main website, sorry i can't give you steps on how to do it cuz i am really really busy, i'll come back later if i can, sorry.

quincy

Date: Jan 18, 2005

actually after i relooked at it, 52 is 3.25% of ______________ is easy, change 3.25 to a fraction, which is 13/4, then flip that and multiply by 52, then times 100 cuz it's a percent. so the answer is 52 x 4/13 x 100

the answer is 1600

quincy

webmaster · Administrator

I'm back and have a little time now.

For the % problems, it is easier to look at it in this fashion. I know you probably already know this but "is" means equals, and "of" means multiply.

So for the first problem : 42% of 27 is _____% of 6% of 25.

Think of it in these terms: .42*27 = x * .06 * 25.

Solving for x we get x = (.42*27)/(.06*25) = 7 * 27/25 = 7 * 1.08 = 7.56

The answer is 756, since we are dealing with percents and not just a regular number. This isn't that hard or time consuming. You should see immediately that the decimals don't matter because they are going to cancel each other out. That leaves only 42/6 * 27/25 which is 7 * 27/25. Not too bad, but I would probably skip this one and come back after finishing the test. So...

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For the second question, read:

http://www.math-magic.com/geometry/determine_tri.htm

But the answer is going to be 16 I believe.

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For the third question, read:

http://www.math-magic.com/trig/triples.htm

But the answer is 181

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For the last question, follow Quincy's instructions

Hope this helps,

Webmaster

-- Edited by webmaster at 11:14, 2005-01-19

-- Edited by webmaster at 11:18, 2005-01-19

vvr1590 · Senior Member

Thanks,

On the first problem, could I leave the answer in the form 189/25?

Also, I looked on the determining triangles page, and I still don't get it. How did you get 16?

Vinay

vvr1590 · Senior Member

if they asked for the largest integral value to make an obtuse triangle, do I just add the two sides and decrease by one?

Vinay

webmaster · Administrator

Vinay,

You can't use 189/25 because this is in decimal format and you are looking for something in percent format which means you have to multiply by 100. So technically you would have 18900/25 which reduces to 756. If it had not said %, you could have used 189/25, but you have to be careful and know what they are looking for.

First off, this next method with obtuse triangles I am a little choppy on as well. This is a trick that came after my time. We know that an obtuse triangle with sides: a, b, and x would have to be between:

|a-b|<x<sqrt(a^2 - b^2) and sqrt(a^2 + b^2)<x<a+b.

Now if a and b are equal, the first section is going to be 0<x<0 which can't happen so we have to focus on the last section. So with a,b = 11 we get sqrt(242) < x < 22. The sqrt(242) is 15.something so x has to be between 16 and 21 inclusive. So the smallest value is 16 and the largest value is 21.

I think I need to update the website, because it didn't take into account the fact a and b could be equal or even possible close enough to where the right or left side would be disregarded.

Thanks for the question. I hadn't thought of that scenario.

Webmaster

vvr1590 · Senior Member

Glad I was able to point out something you hadn't thought of, hehe.

One more question: How do you take arcsin and arccos and arctan of negative values?

Are they the same as the regular angles, only negative?

Thanks

Vinay

-- Edited by vvr1590 at 19:07, 2005-01-20

Date: Jan 20, 2005

you mean something like this?

arccos(-30) is _____?

that's easy, a negative sign just means you go clockwise among the quadrants, so basically, instead of start w/ zero, think of it as 360, and in this case, 360-30=330, and that means the problem could be rewritten like this:

arccos (330) is______?

the answer is sqrt (3) /2

basically what i mean is this...

arccos (-30) = arccos (330) because....-30 degree is the same as 330 degree

hope i explained it to the point where you can understand...i am bad at explaining...

another question: do you know how to find the sum of the first N triangler numbers? (other than just adding them)

problem: the sum of the 1st of 7 triangle number is....

i don't know how to manipulate the numbers to come up w/ a formula

quincy

Date: Jan 21, 2005

hmm...

I can't find a formula, yet...

but here's a list of the sum of the first n triangular numbers, if it helps you to find a formula

1 - 1

2 - 4

3 - 10

4 - 20

5 - 35

6 - 56

7 - 84

8 - 120

Maybe you can use these to find a formula. I'm in biology class right now, so I can't really think about this, but I certainly will keep trying.

Vinay

Date: Jan 21, 2005

The sum of the nth triangular number is n(n + 1)/2. For example : Find the 10th triangular number. Solution : 10(10 + 1)/2 = 55. It is the same formula that is used to find the sum of the first n counting numbers.

Leo Ramirez
www.rammaterials.com

Date: Jan 21, 2005

nonono, mr.ramirez, you misunderstood my question, what i really meant was something like this

the sum of the 1st 3 triangular numbers is _________

to solve this, i need to add the 1st triangular number (1) plus the 2nd (3) then plus the 3rd (6). Therefore, the answer should be 1+3+6, which is 10, and what i am trying to do is find a formula for that instead of adding, the formula you gave us is the formula to find the nth triangular number, but not the SUM of the first nth triangular number, but thanks anyways. :)

quincy

vvr1590 · Senior Member

Thanks for the insight about the counting numbers, Mr. Ramirez, but what if the problem read:

Find the sum of the first 7 triangular numbers. ________

Quincy, is this what you're talking about, or do I have the idea skewed?

Also, my teacher, Mrs. Curry, would be very interested in your Calculator and Numbersense Workshops, as she coaches both UIL contests. How should she get in touch with you?

Vinay

vvr1590 · Senior Member

oops, sorry quincy, I hadn't see your post before posting mine

Date: Jan 21, 2005

lol, yeah, you are right about my meaning, btw, have you been to a contest that has the UIL A set material for this year yet? we're going to Georgetown to compete tomorrow (i can't go cuz i have SATs), and they're using set A material, just wondering if you've taken it and how did you do on it.

quincy

webmaster · Administrator

Alrighty then,

I have something on the sum of triangular numbers, but I will need your help on this. Maybe altogether we can come up with a good formula.

Here is what I came up with so far:

Something you should not is that the sum of consecutive triangular numbers is actually the larger number squared. So for example, the sum of the 5th and 6th triangular numbers is 36 or 6^2. So let's take this idea for working sum of the nth triangular numbers.

1st + 2nd + 3rd + ... + nth = (1st + 2nd) + (3rd + 4th) + ... + ((n-1)th + nth)

*NOTE: First off let's assume n is even.

This will become 2^2 + 4^2 + 6^2 + ... + n^2.

So how do we break this up? Let's factor out a 2^2 or a 4 and get:

4 x [1^2 + 2^2 + 3^2 + ... + (n/2)^2] = 4 x (a)(a+1)(2a+1)/6 with a = n/2 *See sequences

So now we have a basic formula that will work as long as we are looking for an even sum.

Now, what I need your help on is if it is odd. The only thing that I can come up with is finding the sum of the sequences as down above and add in the nth triangular number. So for our example we are going to have the sum of the first 7 triangular numbers. In doing this, we take the sum of the first 6 triangular numbers and get:

4 x 3(4)(7)/6 = 4 x 14 = 56. Now we need to add the 7th triangular number which is 7(8)/2 or 28. So 28 + 56 = 84. I know though this is not the easiest way of doing this. There HAS to be a simpler formula that will work. See what you guys can come up with.

Just to make sure the above method works we can check on the list that is added above by Vinay.

I look forward to seeing more of this discussion. In fact, I am going to make this into a seperate thread.

Webmaster

vvr1590 · Senior Member

Well, I was doing some thinking about this last night, here's what I've got so far:

The sum of the first n triangular numbers can be represented as the sum of:

1
1+2
1+2+3
1+2+3+4
1+2+3+4+5

and so on for as many terms as needed. So if the last triangular number is represented by the nth triangular number, it would then be

1
1+2
1+2+3
1+2+3+4
1+2+3+4+5 ... until the nth triangular number is reached

If we add the columns up, we get n * 1 + 2(n-1) + 3(n-2) .... until the number inside the parentheses becomes one.

So then, that can expand to

an-b

a, is the nth triangular number. i still can't figure out the pattern for b.

For example, for n = 3, we would have

1
1+2
1+2+3

3n + 2(n-1) + 3 (n-2)

which becomes 6n - 8

for n = 4,

it becomes 10n - 20, or 40-20 = 20

- n = 5, then the expression for the sum is 15n - 40, or 75-40 = 35

If you guys could find an expression for the term that is subtracted, then I think we would have the formula that could work for any n.

I'll try to look at your formula, mr. jones, and see how I can help.

Vinay

vvr1590 · Senior Member

Well, I don't know about A material or anything like that, but

we went to a practice meet in November, and I didn't do to well.

I got 5th place, with a score of 125.

But that was before I'd started practicing, so my scores are improving.

Vinay