## Poll

 I love math! 21 votes (45.65%) Math is great. 14 votes (30.43%) My religion is mathology. 6 votes (13.04%) Women didn't speak to me until I was 30. 3 votes (6.52%) Total eclipse reminder -- 04/08/2024 12 votes (26.08%) I steal cutlery from restaurants. 3 votes (6.52%) I should just say what's on my mind. 6 votes (13.04%) Who makes up these awful names for pandas? 5 votes (10.86%) I like to touch my face. 12 votes (26.08%) Pork chops and apple sauce. 10 votes (21.73%)

46 members have voted

charliepatrick
• Posts: 2958
Joined: Jun 17, 2011
September 13th, 2023 at 6:44:04 AM permalink
Quote: ThatDonGuy

...I vaguely remember somebody posting a 40-ticket solution to a 6/49 game that guaranteed at least one number would have at least 3 hits....

After proposing my solution I did stumble across a 163-ticket wheel - http://lottery.merseyworld.com/Wheel/Wheel.html but it's confusing as the next page then looks at 57.
Gialmere
• Posts: 2957
Joined: Nov 26, 2018
September 13th, 2023 at 7:00:03 AM permalink
Newspapers in 1944 noted a striking coincidence:

Is this a bizarre, numerical quirk, or can it be explained?
Have you tried 22 tonight? I said 22.
charliepatrick
• Posts: 2958
Joined: Jun 17, 2011
September 13th, 2023 at 7:41:15 AM permalink
Using the above idea one could divide the range into three parts
(i) 1-14 (ii) 15-28 (iii) 28-37.
Thus given there are five winnings numbers there are either two in (i) or two in (ii) or three in (iii).
The latter can be covered by the lines 28-32 and 33-37 since at least one row must have two.
I've yet to check this but the numbers 1 thru 14 can be covered with 13 lines as follows:-
(it uses the logic used above, have 1-2 with all the remaining numbers 3-14, similarly 3-4 with 5 to 14, then 5-6 with 7 to 14, then do the rest somehow. Mirror logic suggests it could start 1-5 and end 10-14, I don't know )
There might be a way to bring this down to twelve lines, for instance 7 8 appear together a lot, but I don't know.
1 2 3 4 5
1 2 6 7 8
1 2 9 10 11
1 2 12 13 14

3 4 6 7 12
3 4 8 9 13
3 4 10 11 14

5 6 7 8 13
5 6 9 10 14
5 6 8 11 12

7 8 10 11 14
7 9 11 12 13
10 12 13 any any

This give 13 lines to find any pair of numbers in 1-14.

Do the same for 15-28 (you probably only need one or two lines fewer as you don't need to cover No 28 but only 15-27; there may also be a smart way to use 10 12 13 any any etc. across the boundary.)

Thus this currently gives 13 (for 1-14) + 13 (for 15-28) + 2 (for 28-32 and 33-37), giving 28. But I suspect a few more lines can be removed.
charliepatrick
• Posts: 2958
Joined: Jun 17, 2011
Thanked by
September 13th, 2023 at 7:44:56 AM permalink
Quote: Gialmere

Newspapers in 1944 noted a striking coincidence:

Is this a bizarre, numerical quirk, or can it be explained?

It can be explained because the first two numbers (assuming it's 31st December or similar) of your current age and birth year will always add up to the current year. Similarly for something happened (took office) and time since that happened (years in office). And 3888 is twice 1944.
Wizard
• Posts: 26558
Joined: Oct 14, 2009
Thanked by
September 13th, 2023 at 9:15:04 AM permalink
Quote: Gialmere

Newspapers in 1944 noted a striking coincidence:

Is this a bizarre, numerical quirk, or can it be explained?

There are magic tricks based on this principle.

As was pointed out, the first two numbers will sum to 1944 for everybody who was alive in 1944.

The second two will always sum to 1944 for everyone "in office" at that time.

I personally don't do this magic trick because I find it obvious how it's done, but maybe I should try it and see how it goes.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
charliepatrick
• Posts: 2958
Joined: Jun 17, 2011
September 13th, 2023 at 9:15:19 AM permalink
If one splits the range into 1-9, 10-18, 19-27, 28-37 then there are nine numbers in the first three groups and ten in the last. There must be at least two winning numbers in one of the blocks. Using the same logic there are five lines to cover 9 numbers.
1 2 3 5 8
1 2 3 6 7
1 2 3 4 9
4 5 6 7 8
5 6 7 8 9

I've quickly look at the last group and get seven lines
28 29 30 31 32
28 29 33 34 35
28 29 xx 36 37
30 31 32 33 34
30 31 35 36 37
xx 32 35 36 37
33 34 35 36 37

Hence this gives 5+5+5+7=22 lines.

(I haven't double checked this covers everything, but I think four blocks is a reasonable plan of attack.)
aceside
• Posts: 500
Joined: May 14, 2021
September 13th, 2023 at 10:21:53 AM permalink
Quote: Wizard

Quote: Gialmere

Newspapers in 1944 noted a striking coincidence:

Is this a bizarre, numerical quirk, or can it be explained?

There are magic tricks based on this principle.

As was pointed out, the first two numbers will sum to 1944 for everybody who was alive in 1944.

The second two will always sum to 1944 for everyone "in office" at that time.

I personally don't do this magic trick because I find it obvious how it's done, but maybe I should try it and see how it goes.

Good point! Hitler became the chancellor of germany January 30, 1933
Wizard
• Posts: 26558
Joined: Oct 14, 2009
Thanked by
September 13th, 2023 at 12:03:45 PM permalink
Quote: charliepatrick

If one splits the range into 1-9, 10-18, 19-27, 28-37 then there are nine numbers in the first three groups and ten in the last. There must be at least two winning numbers in one of the blocks. Using the same logic there are five lines to cover 9 numbers.
1 2 3 5 8
1 2 3 6 7
1 2 3 4 9
4 5 6 7 8
5 6 7 8 9

I've quickly look at the last group and get seven lines
28 29 30 31 32
28 29 33 34 35
28 29 xx 36 37
30 31 32 33 34
30 31 35 36 37
xx 32 35 36 37
33 34 35 36 37

Hence this gives 5+5+5+7=22 lines.

(I haven't double checked this covers everything, but I think four blocks is a reasonable plan of attack.)

Very good work there. It certainly humbles me. I've gone through and can't find a way to do better than 5 tickets with 9 numbers and 7 with 10.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Gialmere
• Posts: 2957
Joined: Nov 26, 2018
September 13th, 2023 at 5:42:44 PM permalink
Quote: charliepatrick

It can be explained because the first two numbers (assuming it's 31st December or similar) of your current age and birth year will always add up to the current year. Similarly for something happened (took office) and time since that happened (years in office). And 3888 is twice 1944.

Correct!!
Quote: Wizard

There are magic tricks based on this principle.

As was pointed out, the first two numbers will sum to 1944 for everybody who was alive in 1944.

The second two will always sum to 1944 for everyone "in office" at that time.

I personally don't do this magic trick because I find it obvious how it's done, but maybe I should try it and see how it goes.

Correct!!

Good show.

This works for any group of people — as long as they’re all still alive and working, these values will add to a miraculous constant. Here, for example, is the Supreme Court in 2009:

Their magic total, sure enough, is just twice 2009.

------------------------------------------------

We have a picture of him in uniform with two lightning bolts on his helmet.
Have you tried 22 tonight? I said 22.
ThatDonGuy
• Posts: 6349
Joined: Jun 22, 2011
September 13th, 2023 at 6:40:02 PM permalink
Quote: Gialmere

Good show.

This works for any group of people — as long as they’re all still alive and working, these values will add to a miraculous constant. Here, for example, is the Supreme Court in 2009:

Their magic total, sure enough, is just twice 2009.

It only "always" works on December 31. Otherwise, if at least one person in the group has already had their birthday/anniversary that year and at least one has not, the numbers will differ by one.

Gialmere
• Posts: 2957
Joined: Nov 26, 2018
September 14th, 2023 at 7:00:06 AM permalink

Four of the 8 vertices of a cube are vertices of a regular tetrahedron.

Find the ratio of the surface area of the cube to the surface area of the tetrahedron.
Have you tried 22 tonight? I said 22.
ChesterDog
• Posts: 1540
Joined: Jul 26, 2010
Thanked by
September 14th, 2023 at 7:24:37 AM permalink

Gialmere
• Posts: 2957
Joined: Nov 26, 2018
Thanked by
September 14th, 2023 at 5:29:45 PM permalink
Quote: ChesterDog

Correct!!

Well done.

It's neat how platonic solids interact. The way a tetrahedron fits into a cube is a good example. It's no wonder that Kepler thought he was discovering God's master plan when he was working with them.

------------------------------------------------------

Have you tried 22 tonight? I said 22.
Gialmere
• Posts: 2957
Joined: Nov 26, 2018
September 15th, 2023 at 7:00:04 AM permalink

Take the circumference of the earth at the equator to be 24,000 miles. An airplane taking off at the equator and flying west at 1000 miles an hour would land at exactly the same time that it started. Moreover, the sun would not move in the plane’s sky during the flight.

At what degree of latitude could a plane flying 500 miles per hour keep up with the sun in this way?
Have you tried 22 tonight? I said 22.
aceside
• Posts: 500
Joined: May 14, 2021
Thanked by
September 15th, 2023 at 7:38:32 AM permalink
60 degrees
Wizard
• Posts: 26558
Joined: Oct 14, 2009
September 15th, 2023 at 4:15:14 PM permalink
I plan to make a future Ask the Wizard question about the Cash 5 Lottery. Here is a preview of my solution.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Wizard
• Posts: 26558
Joined: Oct 14, 2009
Thanked by
September 15th, 2023 at 4:19:56 PM permalink
Quote: Gialmere

At what degree of latitude could a plane flying 500 miles per hour keep up with the sun in this way?

I get 60 degrees north or south latitude. This is where the circumference is half that of the equator.

As an very unrelated question, how is it that Superman went back in time when he spun around the earth really fast?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Gialmere
• Posts: 2957
Joined: Nov 26, 2018
September 15th, 2023 at 7:26:30 PM permalink
Quote: aceside

60 degrees

Quote: Wizard

I get 60 degrees north or south latitude. This is where the circumference is half that of the equator.

As an very unrelated question, how is it that Superman went back in time when he spun around the earth really fast?

Correct!

Very good.
------------------------------------------

In his book, Sled Driver, SR-71 Blackbird pilot Brian Shul writes:

"I'll always remember a certain radio exchange that occurred one day as Walt (my back-seater) and I were screaming across Southern California 13 miles high. We were monitoring various radio transmissions from other aircraft as we entered Los Angeles airspace. Although they didn't really control us, they did monitor our movement across their scope.

I heard a Cessna ask for a readout of its ground speed."90 knots" Center replied. Moments later, a Twin Beech required the same. "120 knots," Center answered. We weren't the only ones proud of our ground speed that day as almost instantly an F-18 smugly transmitted, "Ah, Center, Dusty 52 requests ground speed readout." There was a slight pause, then the response, "525 knots on the ground, Dusty." Another silent pause.

As I was thinking to myself how ripe a situation this was, I heard a familiar click of a radio transmission coming from my back-seater. It was at that precise moment I realized Walt and I had become a real crew, for we were both thinking in unison. "Center, Aspen 20, you got a ground speed readout for us?" There was a longer than normal pause....

"Aspen, I show 1,742 knots." (That's about 2005 mph)

No further inquiries were heard on that frequency.
Have you tried 22 tonight? I said 22.
MichaelBluejay
• Posts: 1620
Joined: Sep 17, 2010
September 22nd, 2023 at 7:18:06 PM permalink
From the local weekly newspaper here. I posted some of those here before, retyping them, but I finally figured out how to post the pics and link to the solution.

I'm not great with geometry, but even I got this one right (though I did it by a different method than what's described in the answer).

Presidential Election polls and odds: https://2605.me/p
Gialmere
• Posts: 2957
Joined: Nov 26, 2018
October 17th, 2023 at 8:33:32 AM permalink

The revolving portion of a splendid barber pole is a cylinder 4 feet tall and 6 inches in radius. The red stripe is painted in a spiral around the cylinder and makes exactly 8 complete turns around it from bottom to top.

How long is the stripe?

(Ignore its width.)
Have you tried 22 tonight? I said 22.
ChesterDog
• Posts: 1540
Joined: Jul 26, 2010
Thanked by
October 17th, 2023 at 9:46:06 AM permalink

The trick is to make a vertical cut in the white and red material and flatten it out. And then use the Pythagorean theorem.

And then rearrange to get:

Last edited by: ChesterDog on Oct 17, 2023
Dieter
• Posts: 5583
Joined: Jul 23, 2014
Thanked by
October 17th, 2023 at 9:49:41 AM permalink

I get ~97.2π inches.

(I seem to remember something about a rope and a cylinder from maybe 3 years ago?)

The ugly bottle makes me seem more attractive.
May the cards fall in your favor.
Gialmere
• Posts: 2957
Joined: Nov 26, 2018
Thanked by
October 17th, 2023 at 5:08:59 PM permalink
Quote: ChesterDog

The trick is to make a vertical cut in the white and red material and flatten it out. And then use the Pythagorean theorem.

And then rearrange to get:

Quote: Dieter

I get ~97.2π inches.

(I seem to remember something about a rope and a cylinder from maybe 3 years ago?)

The ugly bottle makes me seem more attractive.

Correct!!

Very good.

I've also seen a similar problem where a plumber is wrapping tape around a pipe.

Note, if you're a teacher doing this puzzle you can cut out a paper version of ChesterDog's triangle. Tape the 48" side to a pencil, wrap it around with the red hypotenuse becoming the stripe and then present it as innocent visual aid. If the students have trouble, slowly unravel the right angle triangle and watch their eyes widen.

-----------------------------------

Have you tried 22 tonight? I said 22.
Wizard
• Posts: 26558
Joined: Oct 14, 2009
Thanked by
October 17th, 2023 at 5:30:16 PM permalink
Quote: Gialmere

The revolving portion of a splendid barber pole is a cylinder 4 feet tall and 6 inches in radius. The red stripe is painted in a spiral around the cylinder and makes exactly 8 complete turns around it from bottom to top.

How long is the stripe?

(Ignore its width.)

I get sqrt(48^2 + (96*pi)^2) = 305.388726 inches.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Gialmere
• Posts: 2957
Joined: Nov 26, 2018
October 18th, 2023 at 8:38:56 AM permalink
Here's one you'll often see on SAT tests...

Have you tried 22 tonight? I said 22.
Wizard
• Posts: 26558
Joined: Oct 14, 2009
Thanked by
October 18th, 2023 at 10:56:40 AM permalink
Quote: Gialmere

Here's one you'll often see on SAT tests...

80 degrees
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
ThatDonGuy
• Posts: 6349
Joined: Jun 22, 2011
Thanked by
October 18th, 2023 at 4:07:47 PM permalink
Quote: Gialmere

Here's one you'll often see on SAT tests...

Draw line segment AP
AP, BP, and CP are all radii of the circle, so they have the same length
This means APC and BPC are isosceles triangles; PAC = PCA = 20 degrees, and PAB = PBA = 20 degrees, so APB = 180 - 20 - 20 = 140 and APC = 180 - 20 - 20 = 140, and x = 360 - (140 + 140) = 80 degrees.

ChesterDog
• Posts: 1540
Joined: Jul 26, 2010
Thanked by
October 18th, 2023 at 4:24:17 PM permalink

The sum of the interior angles of a quadrilateral is 360 degrees. ABPC is a quadrilateral.

The angle at A is x/2, and the interior angle at P is 360 - x.

So, x/2 + 20 + 360 - x + 20 = 360

40 - x/2 = 0

x = 80 degrees.
Gialmere
• Posts: 2957
Joined: Nov 26, 2018
October 18th, 2023 at 4:48:47 PM permalink
Quote: Wizard

80 degrees

Quote: ThatDonGuy

Draw line segment AP
AP, BP, and CP are all radii of the circle, so they have the same length
This means APC and BPC are isosceles triangles; PAC = PCA = 20 degrees, and PAB = PBA = 20 degrees, so APB = 180 - 20 - 20 = 140 and APC = 180 - 20 - 20 = 140, and x = 360 - (140 + 140) = 80 degrees.

Quote: ChesterDog

The sum of the interior angles of a quadrilateral is 360 degrees. ABPC is a quadrilateral.

The angle at A is x/2, and the interior angle at P is 360 - x.

So, x/2 + 20 + 360 - x + 20 = 360

40 - x/2 = 0

x = 80 degrees.

Correct!!

Well done.

Evidently many students have trouble with this one. Presumably because the Star Trek logo-ish shape is so awkward. Yet a simple line creates two triangles. Triangles have rules that must be obeyed. It's a downhill hike from there.

--------------------------------------------------

Geometry

Just sum.
Have you tried 22 tonight? I said 22.
Wizard
• Posts: 26558
Joined: Oct 14, 2009
October 18th, 2023 at 5:06:10 PM permalink
Quote: Gialmere

Geometry

*groan*

Remind me to tell you my rope joke if we ever meet. It's too long to type out.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Ace2
• Posts: 2691
Joined: Oct 2, 2017
October 18th, 2023 at 5:10:29 PM permalink
Quote: Gialmere

[

Geometry

Just sum.

Best two jokes you’ve posted in a while!!
It’s all about making that GTA
Jimmy2Times
• Posts: 58
Joined: Nov 28, 2018
October 19th, 2023 at 2:30:22 AM permalink
Quote: ChesterDog

The sum of the interior angles of a quadrilateral is 360 degrees. ABPC is a quadrilateral.

The angle at A is x/2, and the interior angle at P is 360 - x.

So, x/2 + 20 + 360 - x + 20 = 360

40 - x/2 = 0

x = 80 degrees.

Why is the angle at A equal to x/2? Thanks
I 'm gonna go get the papers, get the papers.
ChesterDog
• Posts: 1540
Joined: Jul 26, 2010
Thanked by
October 19th, 2023 at 4:05:28 AM permalink
Quote: Jimmy2Times

Quote: ChesterDog

The sum of the interior angles of a quadrilateral is 360 degrees. ABPC is a quadrilateral.

The angle at A is x/2, and the interior angle at P is 360 - x.

So, x/2 + 20 + 360 - x + 20 = 360

40 - x/2 = 0

x = 80 degrees.

Why is the angle at A equal to x/2? Thanks

I used the famous geometry theorem that the measure of an inscribed angle is half the central angle.

Below demonstrates why this theorem is true.

Triangle APB is an isosceles triangle because AP and BP are both radii of the circle. Therefore, angle BAP = angle ABP (∠BAP = ∠ABP).

The angles of a triangle add up to 180 degrees, so ∠BAP + ∠ABP + ∠APB = 180.

∠APD is 180 degrees, so ∠APB = 180 - x/2.

Combining these last three equations yields: ∠BAP + ∠BAP + (180 - x/2) = 180.

Solving yields: ∠BAP = x/4. Therefore, ∠BAC = 2( x/4 ) = x/2.
Gialmere
• Posts: 2957
Joined: Nov 26, 2018
October 23rd, 2023 at 7:55:21 AM permalink
A quick Easy Monday puzzle...

At what time, exactly, do the two hands of a clock overlap for the first time after 12:00 midnight?
Have you tried 22 tonight? I said 22.
Wizard
• Posts: 26558
Joined: Oct 14, 2009
October 23rd, 2023 at 8:32:35 AM permalink
Quote: Gialmere

At what time, exactly, do the two hands of a clock overlap for the first time after 12:00 midnight?

1:05:27 AM.

In other words 12/11 hours.

I solved it by equating the time of both hands when they meet. Remember time = distance/rate. Let d be the distance from the 12:00 position, where each hour is 1/12, then:

minute hand time = hour hand time
(1+d)/12 = d
1+d = 12d
d = 1/11
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Gialmere
• Posts: 2957
Joined: Nov 26, 2018
October 23rd, 2023 at 8:36:58 AM permalink
Have you tried 22 tonight? I said 22.
Wizard
• Posts: 26558
Joined: Oct 14, 2009
Thanked by
October 23rd, 2023 at 10:12:45 AM permalink
Quote: Gialmere

Oy.

I thought the 12/11 hours was an exact expression, but to put it another way 1 hour, 5 minutes, 27 3/11 seconds.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Gialmere
• Posts: 2957
Joined: Nov 26, 2018
October 23rd, 2023 at 4:57:09 PM permalink
Quote: Wizard

Oy.

I thought the 12/11 hours was an exact expression, but to put it another way 1 hour, 5 minutes, 27 3/11 seconds.
[/spoiler]

Correct!!

Very good.

(Apologies for my laziness. I have the flu.)
----------------------------------

You get mad at me when I work....You get mad at me when I don't work.

Sincerely,

Confused alarm clock.
Have you tried 22 tonight? I said 22.
Gialmere
• Posts: 2957
Joined: Nov 26, 2018
October 25th, 2023 at 7:00:44 AM permalink

Mike enters the Vermilion game with a batting average of .274. After going 3-for-4 in the game, his average has gone up to .289.

How many hits did he have for the season before the game started?
Have you tried 22 tonight? I said 22.
unJon

• Posts: 4626
Joined: Jul 1, 2018
Thanked by
October 25th, 2023 at 7:48:28 AM permalink
Quote: Gialmere

Mike enters the Vermilion game with a batting average of .274. After going 3-for-4 in the game, his average has gone up to .289.

How many hits did he have for the season before the game started?

Solving the two simultaneous equations:

Hits/ABs = 0.274

(Hits + 3) / (ABs + 4) = 0.289

Gives hits of 33.683

With rounding you get the right answer with 34 Hits and 124 At Bats before the current game.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Wizard
• Posts: 26558
Joined: Oct 14, 2009
October 25th, 2023 at 8:34:37 AM permalink

There are two houses, A and C, near a straight river. The closest point to the river from A is point B. The closest point to the river from C is D.

AB = 2
CD = 3
BD = 5

It is desired to build a single pumping station along the river to supply water to both houses. The goal is to minimize the piping required to connect both houses to the pumping station. Where should the pumping station be built and how much piping is required if built there?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
DogHand

• Posts: 1565
Joined: Sep 24, 2011
October 25th, 2023 at 9:18:49 AM permalink
Quote: Wizard

There are two houses, A and C, near a straight river. The closest point to the river from A is point B. The closest point to the river from C is D.

AB = 2
CD = 3
BD = 5

It is desired to build a single pumping station along the river to supply water to both houses. The goal is to minimize the piping required to connect both houses to the pumping station. Where should the pumping station be built and how much piping is required if built there?

Wiz,

I get a location on the river 2 units from B and 3 units from D.

The total length is then 2√2 + 3√2 = 5√2.

Dog Hand
chevy
• Posts: 146
Joined: Apr 15, 2011
October 25th, 2023 at 9:19:52 AM permalink

Do you mean that both house are connected directly to the pumping station? If so then I get

****WRONG - can't multiply numbers apparently****
pumping station 1 unit to right of B
pipe required is 5+sqrt(5)=7.236

But building station at B and pipes BA + AC = 2+sqrt(26)=7.099 seems better ......are there others better still?

******EDITED - I agree with Doghand now.....but still unsure it is in fact best??

EDITED mine - I agree with Doghand....but still unsure of actual question.
chevy
• Posts: 146
Joined: Apr 15, 2011
October 25th, 2023 at 9:32:02 AM permalink
Quote: DogHand

Wiz,

I get a location on the river 2 units from B and 3 units from D.

The total length is then 2√3 + 3√2.

Dog Hand

I agree with Dog Hand location, 2 units from B....

I think pipe required is 2sqrt(2) + 3sqrt(2) = 5sqrt(2) = 7.071????

DogHand

• Posts: 1565
Joined: Sep 24, 2011
October 25th, 2023 at 9:34:40 AM permalink
Quote: chevy

Quote: DogHand

Wiz,

I get a location on the river 2 units from B and 3 units from D.

The total length is then 2√3 + 3√2.

Dog Hand

I agree with Dog Hand location, 2 units from B....

I think pipe required is 2sqrt(2) + 3sqrt(2) = 5sqrt(2) = 7.071????

Yeah... I fixed my typo.

I agree with your total length.

Dog Hand
chevy
• Posts: 146
Joined: Apr 15, 2011
October 25th, 2023 at 9:56:42 AM permalink

Starting with Dog Hand solution (shown in black) and dropping perpendicular AF + EC.....Green < Black
But then starting with Green, and Dropping perpendicular FG +AF + FC..... PINK < Green

Not optimizing the locations, just showing that the two direct connections can be improved. Not sure how to optimize if location on the river is a variable.

unJon

• Posts: 4626
Joined: Jul 1, 2018
October 25th, 2023 at 10:07:42 AM permalink
Quote: chevy

Starting with Dog Hand solution (shown in black) and dropping perpendicular AF + EC.....Green < Black
But then starting with Green, and Dropping perpendicular FG +AF + FC..... PINK < Green

Not optimizing the locations, just showing that the two direct connections can be improved. Not sure how to optimize if location on the river is a variable.

Oh that’s an interesting question. Are you supposed to make 3 lengths of pipe if equivalent length?
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
chevy
• Posts: 146
Joined: Apr 15, 2011
October 25th, 2023 at 10:50:43 AM permalink
Quote: unJon

Quote: chevy

Starting with Dog Hand solution (shown in black) and dropping perpendicular AF + EC.....Green < Black
But then starting with Green, and Dropping perpendicular FG +AF + FC..... PINK < Green

Not optimizing the locations, just showing that the two direct connections can be improved. Not sure how to optimize if location on the river is a variable.

Oh that’s an interesting question. Are you supposed to make 3 lengths of pipe if equivalent length?

Not sure of correct answer ...but I don't think you can require equal lengths in general.

Call F = the floating intersection....if all three are equal length, then AF = CF and F must be on the line perpendicular to AC.
But imagine stretching the problem horizontally (BD increases to a large amount)...AF and FC increase
The pipe from F to G (The location on the river)will not grow without limit) in fact FG would never be more than 2.5 units (midpoint between heights of A and C)

So for arbitrary case, seems can't require equal lengths.

Need some constraint though, it seems?

If intersection F is at (x,y), the total of the three pipes would be

AF + FC + FG
sqrt(x^2 + (2-y)^2) + sqrt ( (5-x)^2 + (3-y)^2) + y

???

Wizard