BMAD

My only question is....did it trick the camera too?

Mike joins a campfire with three friends. After five minutes, Mike is bored and decides to start up a conversation: (speaker- dialogue) Mike- Tell me something interesting. Adam-one of us is always a liar. Mike-Really? Eric- Yes. They always lie, but one of us is also always a truth teller. Mike- Oh, cool, well then …that would mean that Sam is the ‘always liar.’ Sam- Yes, yes I am. Mike-…What? Wait a minute. I thought you were always a liar. Sam- Yes I am. Eric-He is. Mike-He is? Eric-yes Sam-yes Mike-yes? Adam-no. Mike-no. Sam-no. Mike-wait….then does that make Eric the liar? Eric- Yes Sam- No Adam-yes Mike- What?? **Who is the ‘always’ liar?

the first student used a 30 degree angle and the second student used a 45 degree angle. The op did not give the third person's angle. I guess I don't understand. Was that not the question? maybe i am confused. but the way i read your solution is that you assumed the first person was 30 degrees, and the third person was 45 degrees and then found the second person. The op gave the first person's angle and the second person's angle and asked for the thirds.

The answer is even closer so if no one can find an even more precise answer then this will be the answer. And before I get some guessess with a bunch of nines, the answer that i found is not consisting solely of running nines after the decimal.

the first student used a 30 degree angle and the second student used a 45 degree angle. The op did not give the third person's angle.

I got the same answer but the rest of the board didn't agree with me.

Not sure of the title of this paradox so I didn't know how to search the forums

I am extremely tempted to mark your answer solved. It is very close. I never looked at the approach like that before for this problem. It gives a very very good approximation one that you have to go out about 8 decimal places to see a difference. I will wait and see if someone can find the more precise answer.

Suppose two cameras were installed x distance apart from each other in a hallway. They were placed directly across from each other in circular casings. The first camera is in a small casing that is about 3 times as small (1/3 the radius) as the bigger casing for the other camera which limits its viewing angle. The distance between the two cameras is such that the max they can see horzontally is the widest point on the other camera's casing. If you were to draw lines of the max viewing area from each camera to the casings of the other camera, the lines would intersect their own casings. Which camera would have the greater distance between its own intersections (intersection to intersection) at its casings.

A standard pack of cards is thrown into the air in such a way that each card, independently, is equally likely to land face up or face down. The total value of the cards which landed face up is then calculated. (Card values are assigned as follows: Ace=1, 2=2, ... , 10=10, Jack=11, Queen=12, King=13. There are no jokers.) What is the probability that the total value is divisible by 13?

Point P is inside ABC, and is such that PAC = 18°, PCA = 57°, PAB = 24°, and PBA = 27°. Show that ABC is isosceles.

Three students are outside looking at the sun during an eclispe (don't worry, they are taking precautions). Assume the students are the same height. The three students are collinear but an unknown distance apart. The first student is able to see the sun tilting up their head at a 30 degree angle. The second student needs to tilt their head at a 45 degree angle in order to see the sun. Now though we don't know how far apart the three students are from each other, we do know that second student is perfectly in the middle between the first and third. What angle must the third student use to see the sun?

In a small community there are four households. The Abuleto Family, Boggs Family, and Carlos Family live are the corners of this community. The Douglas’s live between the Abuleto family and the Bogg’s family but not equidistant between them. There is a dirt path that runs directly from the back of Carlos family and Douglas families’ homes. This dirt path is straight. The neighborhood is surrounded by sidewalks that go to each adjacent house (straight line). The angle between the dirt path from Douglas to Carlos and the sidewalk for Bogg’s and Carlos is fifteen degrees. The distance of the sidewalk between Bogg’s and Carlos is sqrt(6) miles, while the distance between the Abuleto and Douglas families is only 1 mile. What is the angle between the sidewalks of the Abuleto family?

which is 'they' in this context...the object passing through or the object that is being passed through, or both?

if THERE ARE TWO ERRORS IN THE THE TITLE OF THIS PUZZLE is the title then what do we call: "Two wrongs might be OK?" The more i talk on here the more i realize that my english is still not right.

I see what I did wrong. I have it being quarter circles at the end of radius 30.

stake

Can you explain how you found it to be 4100? I found it to be 3425

I'll retire the topic.

Your earlier solution only resolved if the house was indeed a rectangle. As shown in prior posting, giving two dimensions does not necessitate a rectangle. If the house was a different shape (still quadrilateral) what would happen to the max and min possible reach of the dog (tethered at a single point) as we try greater and lesser angles? Sorry Prime, I'm not that creative.

fair enough. by classic, i only meant that I heard it long ago and assumed that it was somewhere buried in the forums (but could not be found by me). It's your puzzle, so your interpretation prevails. However, you mentioned that it's a 'classic' and the prisoners are in a circle, which, in my opinion, strongly suggests the classic interpretation of the problem known as the Josephus problem. The reason they are in a circle is that you don't stop at the end and start the next round with the 'next first person' as if they were in a row. That distinction is what makes this problem interesting, IMHO.

Though you know my feelings towards CAD

This op unlike part 1 states in the last line that it is a straight line. Lines go on forever.

I think we are interpreting the OP differently. In your case, as I read your answer, the executioner never stops, cutting in a circle until they are done. The way I meant for the op to be interpreted is that after the first run of kills, he/she starts back over and kills the new first person, continue through and starting over once they come to the end with the 'next' first person. so with 13 people:round 1,, 1,3,5,7,9,11,13...round 2: 2,6,10...round 3: 4, 12...8 lives I like your solution too though. correct answers Respectfully disagree. Let's test this solution for 5 prisoners. Execution order will be 1,3,5,4 with #2 surviving. For 13 prisoners: 1,3,5,7,9,11,13,4,8,12,6,2 with #10 survivng.