Puzzles

[ByteInBits]

@upsidedown — Bad news 🙁

For now, you get to take the consolation prize, sorry.

Try working your answer backward you need a series of integers
the last of which should be what the five contestants collectively divided amongst themselves.
( There should be no need for any ’rounding’ )

[upsidedown]

Thanks for the consolation prize 🙂

I think went wrong by falsely assuming that the 5 players need to use one of the coins for the door when they go in together, but the puzzle doesn’t say that!

If I change np + 1 to np, my previously useless equation from before gives me 3121, which I now believe to be the correct answer (hopefully).

c₀ = (np + (1 – rᵖ)r / (1 – r)) / rᵖ = 3121

For the extra questions: (1) 2096; (2) 15621 (10500 go in the hole, leaving 1023 per player)

[ByteInBits]

@upsidedown
This late response is due to the Panel’s debate on whether to
award the prize after a first wrong answer was given.

🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂

You will now be pleased to know that your corrected answer makes
you and your team winners.

This was decided due to the fact that you answered ALL questions correctly!

Guards will be assinged to each of you to assist with your safe journey to your bank.

🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂 🙂

[F6EXB_the_frenchy]

@ByteInBits

With the help of Excel:
The division gives 204 coins to each person, so there were 1,020 coins left when the fifth person left. That’s 4/5 of the number of coins there were when he entered + the coin from the door. So there were 1,276 coins when he entered, and he put 255 coins in the hole.
We calculate the same way for the other players.
There were initially 3,121 coins, 5 went to the door and 2,096 into the hole.
The mathematician wins 175 coins, or £17,500. The other players keep 169 coins, so they don’t have £17,000.
I haven’t thought about the last question yet.

[ByteInBits]

F6EXB_the_frenchy

Good answers, there was a correction to the winnings following the post – each got £17,000
(discovering the typo just after posting really annoyed me 😉 )

[Kingswinford-Warriors-Alumni]

@F6EXB_the_frenchy
#112088

(Sorry it took a while to get back to you! I am quite a busy person…)

The triangle is indeed ‘special’ when the median is 4. (As a cheeky bonus bonus question, what if AM=1 instead?)

For the bonus question, Heron’s formula might be needed for the general case, but there is no need for any trigonometry/etc. (There is a very nice, purely geometric solution. See if you can find it!) I therefore FORBID THE USE OF TRIGONOMETRY FOR THE GENERAL CASE! I also deliberately set AM=2 to ensure that Heron’s formula is not needed (you shouldn’t need anything more than the normal area of a triangle formula if you do the reasoning correctly), so I also FORBID THE USE OF HERON’S FORMULA FOR THE BONUS QUESTION!

Good luck.

[ByteInBits]

@Kingswinford-Warriors-Alumni – Post #112062

I think the area of triangle ABC
with AM = 4 is 0 square units (a degenerate triangle?).
For AM = 2, the area is 6 square units.

[Robb27]

@HARRY ‘NET LOSS’ in the News.
I believe Alice and Bob played the first match. Claire then had to have lost the second match in order for Alice to have played 17 games, Bob 15 and Claire 10. If I am also correct, from that point, there are 211 different ways the remaining games could have played out to give those same totals.

Just for fun 17+15+10= 42 / 2 players = 21 games played. With three different starting pairs 3*(1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 + 2048 + 4096 + 8192 + 16384 + 32768 + 65536 + 131072 + 262144 + 524288 + 1048576) = 6,291,453 possible game combinations.

[Kingswinford-Warriors-Alumni]

@ByteInBits
#112104

Correct! Nice work. (Out of interest, what method did you use?)

Some more bonus questions related to my original post (#112062):

1. Calculate ∠BAM when AM=2.
2. Calculate the area of triangle ABC if AM=1 instead. (First posted in response to @F6EXB_the_frenchy in #112176)
3. What happens if AM=0.5?
4. Find a general formula for the area of ABC in terms of AB, AC and AM. Using Heron’s formula is NOW PERMITTED.
5. Come up with a geometric argument for the previous questions! (Abusing trigonometry does work, but can get slightly messy.)

Have fun!

[Note to Harry: I will not post solutions for these questions. The general argument is along the same lines as the solutions I posted for the original question.]

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