Solve Sticky Spitballs Art Puzzle: Bob, Cal, Don & Baum

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  • Thread starter Thread starter Wilmer
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Discussion Overview

The discussion revolves around a logic puzzle involving four individuals (Art, Bob, Cal, and Don) and their respective ages and positions related to sticky spitballs they shot at a wall. Participants explore the relationships between their ages, the positioning of the spitballs, and the clues provided to deduce specific information about the oldest shooter and distances between spitballs.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Participants discuss the arrangement of spitballs based on the clues given, such as Don's spitball being directly right of the RSB and below Baum's spitball.
  • There are conditions regarding the ages of the shooters, with some participants noting that Bob is younger than the RSB shooter and Daly is younger than the WSB shooter.
  • Some participants propose different interpretations of the clues, leading to varying conclusions about the identities of the shooters and their respective ages.
  • One participant expresses agreement with another's solution but notes that their presentation is more visually appealing, suggesting a focus on clarity in problem-solving.

Areas of Agreement / Disagreement

The discussion contains multiple competing views regarding the arrangement of spitballs and the ages of the shooters. No consensus is reached on the final answers to the puzzle.

Contextual Notes

The puzzle relies heavily on the interpretation of the clues provided, and participants may have differing assumptions about the relationships between the ages and positions of the spitballs. Some mathematical steps remain unresolved, contributing to the ongoing debate.

Wilmer
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Art, Bob, Cal and Don (ages and last names not necessarily in order:
31 32 33 34, Baum Baxter Daly Neary) each decide to shoot a sticky
spitball at the wall across the beverage room.

To permit this (else this puzzle could not be devised and you'd all
be disappointed) 4 spitballs magically appear on their table:
a Blue SpitBall (BSB), a green one (GSB), a red one (RSB) and
a white one (WSB).

OK: the shooting's done; believe it or not, here's how the SB's stuck:
-Don's SB directly right of the RSB and directly below Baum's SB
-Neary's directly below RSB, Don's 72 inches from Baxter's
-Art's 97 inches from Neary's and 35 inches from GSB

It is entirely possible that you'll need these clues:
-Bob is younger than the RSB shooter, Daly younger than WSB shooter
-Cal is older than BSB shooter, Art older than Baxter

What is the first name of the oldest shooter,
the distance in inches from Bob's SB to the RSB,
and from Daly's SB to the BSB?
 
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In case I forget it!
Interpreting the clues properly will force Art as the oldest,
and place the spitballs a bit like this:
Code: 
                                             sb1                                             
                    sb2                      sb3                    sb4
...which gives you two 90 degree triangles:
(sb1 to sb3) = 35, base = 72, (sb1 to sb4) = 97.
Solve to get (sb2 to sb4) = 30, (sb3 to sb4) = 78.
sb1: Art Baum, white
sb2: Cal Baxter, red
sb3: Don Daly, green
sb4: Bob Neary, blue
 
Wilmer said:
Art, Bob, Cal and Don (ages and last names not necessarily in order:
31 32 33 34, Baum Baxter Daly Neary) each decide to shoot a sticky
spitball at the wall across the beverage room.

To permit this (else this puzzle could not be devised and you'd all
be disappointed) 4 spitballs magically appear on their table:
a Blue SpitBall (BSB), a green one (GSB), a red one (RSB) and
a white one (WSB).

OK: the shooting's done; believe it or not, here's how the SB's stuck:
-Don's SB directly right of the RSB and directly below Baum's SB
-Neary's directly below RSB, Don's 72 inches from Baxter's
-Art's 97 inches from Neary's and 35 inches from GSB

It is entirely possible that you'll need these clues:
-Bob is younger than the RSB shooter, Daly younger than WSB shooter
-Cal is older than BSB shooter, Art older than Baxter

What is the first name of the oldest shooter,
the distance in inches from Bob's SB to the RSB,
and from Daly's SB to the BSB?

From the clues, the configuration of the SBs on the wall would appear to be the following:
$$\begin{array}{llll} {} & {} & {} & \bullet\ \text{Baum} \\ {} & {} & {} & {} \\ \bullet\ \text{RSB} & {} & {}
& \bullet\ \text{Don} \\ {} & {} & {} & {} \\ \bullet\ \text{Neary} & {} & {}
& {} \end{array}$$
So Don is not Baum or Neary; he’s not Baxter either, whose SB is 72 inches away. Therefore Don is Daly, and Baxter’s is the RSB.

Art is not Neary (whose SB is 97 inches away) so he’s either the RSB shooter or Baum. But he can’t be the former since we now know that the RSB shooter is Baxter and Art is older than Baxter. Therefore Art is Baum.

This leaves Bob and Cal as Baxter and Neary. Bob is not the RSB shooter, who is older than Bob and who we know is Baxter; therefore Bob is Neary, leaving Cal as Baxter.

The oldest shooter can’t be Bob Neary or Don Daly, each of whom is younger than somebody else; nor is it Cal Baxter, than whom Art Baum is older. Therefore $\fbox{Art is the oldest}$.

Now, some geometry. Art Baum’s SB is 97 inches from Bob Neary’s, and 35 inches from the GSB; looking at the diagram, the GSB has got to be Don Daly’s. Let the distance between Neary’s SB and the RSB be $x$ inches; then Pythagoras’s theorem gives
$$(35+x)^2+72^2\ =\ 97^2$$
$\implies\ x\ =\ 30$.

So Bob’s SB is $\fbox{30 inches}$ from the RSB.

Finally, we work out who shot the BSB. He is younger than Cal Baxter and so can’t be Cal Baxter – or Art Baum, who we now know is the oldest. We also now know that Don Daly shot the GSB. Therefore it was Bob Neary. By Pythagoras again, his SB is $\sqrt{30^2+72^2}\ =\ \fbox{78 inches}$ from Daly’s SB.
 
YES! Your solution is kinda same as mine;
BUT "looks" much better!