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Could someone beneficially confirm this? Thanks.

Never occurs purely in nature, starts with L.

**Mathegocart**- Replies: 2

I have had a problem with this perplexing problem: The triangle ACB is an isosceles triangle.

AC = AP = PQ = QB

So I though that since ACB is isosceles, we can conclude that 180-6x+2x=3x. 180-4x=3x 180=7x, x = 180/7.

Confirmation needed

bobbym wrote:

Think about it for a bit. I expect that you can get it on your own.

If you get stuck take a look at all the possibilities that can happen.

{5, 5, 5, 1, 1}

{5, 5, 1, 5, 1}

{5, 5, 1, 1, 5}

{5, 1, 5, 5, 1}

{5, 1, 5, 1, 5}

{5, 1, 1, 5, 5}

{1, 5, 5, 5, 1}

{1, 5, 5, 1, 5}

{1, 5, 1, 5, 5}

{1, 1, 5, 5, 5}What do you notice when the 3 fives run out first?

Then there are solely pennies left.

bobbym wrote:

Hi;

Hmmm, I will try to find a more intuitive way.. I felt like it was so simplistic I dismissed it as trivial..

bobbym wrote:

Urn, bag, box, bucket... problems are never simplistic.

Could you confirm that?

Also, I have another problem.

3) There are 3 nickels and 2 pennies in a bag. A man draws coins out of the bag without replacement. He stops when all 3 nickles or 2 pennies are drawn out. What is the probability that he pulls 3 nickels?

A: Obviously this is quite simplistic: 3/5*2/4*1/3 = 1/10, though the answer choices do not concur(there was no 1/10 as a choice.). Am I simply missing an obvious step? I want some confirmation.

bobbym wrote:

Hi;

ARGH! I had attained that solution though I doubted myself and picked 44.

**Mathegocart**- Replies: 15

Today I took a intriguing mathematical quiz(cannot disclose: sorry!)

Anyhow I feel like it devastated me, and I have two questions that were perplexing, namely..

1) Andy and Barry are running around a circular track that is 400 ft. Andy and Barry start at the same place, though Barry pulls up, travelling at 125% the speed of Andy. How many laps of the track has he traversed once Barry surpasses Andy?

2) There are 216 sprinters participating in a race. Unfortunately, the track only holds 6 people at a time. Once a sprinter wins the race, the others are eliminated. This process is reiterated until a final champion wins. How many races are held to find the champion?

bobbym wrote:

That is true. It always has been a big mystery to me why people prefer the rudeness of the AOPS and flock to that place. Sure, they have more help but that is only because so many have gone there.

I have scoured the AOPS forums and while it seems a little more informal, it seems to have little insolence.

-MG

Zeeshan 01 wrote:

sin = opp/hyp what is opp i konw only base altitude hypotenus

Opposite is the side *opposite* the angle specified, adjacent is the side that is consecutive with the angle, and you know what hyp is.

I have found my password!

Ναί!

**Mathegocart**- Replies: 1

I have been bombarded with a particularly perplexing logic problem which states as follows.

Middle, Rear and Front are standing in a line. A eccentric millionaire challenges them to a game. Choosing arbitrarily from 3 white hats and 2 black hats, the millionaire puts a hat on each of the three men. The men cannot see their own hats. The millionaire then asks each of the three the color of their hats. Rear, who can see two other hats, declines to give an answer(he cannot determine his hat's colour.). Middle, who can see only Front's hat, also declines to give an answer(he cannot determine his hat's colour.). Finally, Front, who cannot see any hats, states the color of his own hat.

How?

If somebody decides to link me to the MIF "Three Hats Puzzle", keep in mind we have THREE white hats and TWO black ones.

Now I disprove that scenarios with Front having a black hat could happen in these conditions(Either Middle or Rear would be able to clearly define the colour of their hat.), and thus Front has a white hat. Wikipedia concurs.

However, I have noticed that I end with a convoluted, case by case solution. Can anyone provide me with a simpler solution(not case by case..)?

Zeeshan 01 wrote:

Ok but how this If there is -x how we go to this step -xlne and second step is lne^-x ln(c.e^-x)

That is a property of logarithms. 2log(10) = log(10^2).

Could you please clarify your last step? What is c.e^x????

I am happy I could incite and spark debate about the enigma, the Earth and the String.

**Mathegocart**- Replies: 1

Hello fellow genial friends,

in the diagram , is < 1 = < 3 and <2 = <4?

**Mathegocart**- Replies: 1

1. There are three people, A,B, and C. They are waiting in a barren place where a person asks which of the three has the treasure.

A: Not me

B: I have it

C: Not B.

Strangely, 2 of them always lie, and only 1 tells the fruitful truth.

It is obviously A through casework, but is there another way?

bob bundy wrote:

hi Mathegocart and thickhead

I've searched their site but cannot find any dimensions for their trucks. Strange! Wouild you buy a vehicle without knowing how big it is? Pictures with a trucker in shot suggest they are only about 12 ft tall. If a truck really was 17 ft tall it wouldn't fit under UK motorway bridges.

Bob

ps. LATER EDIT: This is how to cut off the main route from the rest of England to the Dover continental ferry port and the Channel Tunnel:

What a calamity! They look as if to fit under most US bridges here..

Correct? Incorrect?

**Mathegocart**- Replies: 38

A string is wrapper around the Earth's equator(i.e, the circumference) and the two ends of the string just touch. Now suppose that another string is tied to the original string so that it is 100 ft longer. If this new string is placed around the equator and pulled tight so it is suspended over the earth, how high will the string be above the ground?

a. An atom high

b. A bowling ball high

c. A Mack truck high

d. Exactly 100 ft.

I believe it is C.

Explanation:

Let CE = Earth's circumference and let CS = String's new circumference

Let RE= Radius of the Earth and let RS = Radius of the string

CE = 2πRE

and CR+100 = 2πRS.

We want to find the difference between the two radiuses(aka height.)

C/2π = RE

C+100/2π= RS

thus...

RS - RE = (C+100-C)/2π.

RS - RE = 100/2π = 50/π

50/pi ≈ 17.

Therefore Mack truck, aka C, is the solution.

I would LaTeX the aforementioned equations but I'm in a hurry.

bobbym wrote:

Hi;

I am getting something around 46..

**Mathegocart**- Replies: 5

has been given to me as homework to be done. It asks for me to give a proper explanation why as well. Optimal path refers to the path that would minimize the length of the voracious ant's travel. I am thoroughly befuddled by the "it includes the floor and ceiling" hunt as the optimal path does not seem to go through as presupposed.

thickhead wrote:

Verified, correct.

bobbym wrote:

Hi;

Correct, verified.