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GROUPS:
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3
John Custy
Notes on a Car Talk Puzzler
John Custy, Electrical Engineer (retired)
Posted
3 months ago
258 Views

0 Replies

3 Total Likes
Follow this post

On the 14 September 2020 episode of
C
a
r
T
a
l
k
, Tom and Ray posed the following puzzler. (A transcript, available
h
e
r
e
, is well worth reading). I like this puzzler because the proposed answer, though incorrect, has a certain playful appeal. It also gives an excuse to play with Integrate, Region, and ProbabilityDistribution.
The fuel tank of a truck has the shape of a cylinder on it’s side, with a filler hole on top. When the tank is half full of fuel, a broomstick inserted through the filler opening will be marked at half the tanks’s diameter. Where is the stick marked when the tank is 1/4 full?
The proposed answer involves a circular cardboard disk, as typically found under a pizza. The circular disk is sliced precisely in half, and the point where the halfdisk balances on a pencil point is found. This point is said to indicate the fuel level when the tank is 1/4 full.
This figure will help clarify ideas and notation.
P
l
o
t
[
{
1

S
q
r
t
[
1

2
x
]
,
C
a
l
l
o
u
t
[
I
f
[
1

S
q
r
t
[
1

2
x
]
<
0
.
4
5
,
0
.
4
5
,
N
o
n
e
]
,
"
F
u
e
l
L
e
v
e
l
h
"
,
S
c
a
l
e
d
[
0
.
2
]
]
}
,
{
x
,

1
,
1
}
,
E
p
i
l
o
g
{
T
e
x
t
[
"
A
r
e
a
B
e
n
e
a
t
h
\
n
C
h
o
r
d
(
A
B
C
)
"
,
{
0
.
3
,
0
.
3
}
]
}
,
F
i
l
l
i
n
g
{
1
{
2
}
}
,
I
m
a
g
e
S
i
z
e
4
0
0
,
F
r
a
m
e
T
r
u
e
,
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r
a
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e
L
a
b
e
l
{
x
,
R
o
t
a
t
e
[
y
,

π
/
2
]
}
,
P
l
o
t
L
a
b
e
l
"
C
r
o
s
s
s
e
c
t
i
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n
o
f
f
u
e
l
i
n
a
c
y
l
i
n
d
r
i
c
a
l
t
a
n
k
(
p
a
r
t
i
a
l
v
i
e
w
)
\
n
\
t
B
l
u
e
c
u
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e
i
s
y
=
1

S
q
r
t
[
1

x
^
2
]
"
]
O
u
t
[
]
=
The figure represents half the fuel tank along with some fuel; clearly, a stick inserted vertically through the filler cap will be wetted to a level
h
. The shaded twodimensional area serves as a proxy for fuel volume, and we’ll attempt to reduce verbiage by referring to it as the Area Beneath Chord, or ABC. Our unit circle has total area
π
2
r
=
π
, and so
A
B
C
=
π
when
h
=
2
,
A
B
C
=
π
/
2
when
h
=
1
, etc. The challenge is to find
h
that makes
A
B
C
=
π
/
4
.
The point along
x
=
0
on which a cardboard semicircle will balance is proposed as the correct value of
h
. This is plausible: after all, a cardboard shape balanced on a point has “equal stuff” all around, right? Unfortunately, what’s equal is the torques, and not necessarily the masses (though this approach would work for a rectangular tank). A remedy is possible with a balance scale: the cardboard can be sliced along a candidate fuel level, and the problem is solved when the two pieces put the scale in equilibrium. The only disadvantage is that trial and error, and perhaps a lot of cardboard, is required.
Fortunately, cardboardfree solutions are possible.
A Solution using Integrate
Our fuel tank is represented by a unit circle centered at (0,1). We’ll need
x
as a function of
y
for Integrate, and again below in ProbabilityDistribution.
S
o
l
v
e
[
2
x
+
2
(
y

1
)
1
,
x
]
x

2
y

2
y
,
x
2
y

2
y
O
u
t
[
]
=
The amount of fuel in the tank is related to
h
as follows.
C
l
e
a
r
[
a
b
c
]
;
a
b
c
=
I
n
t
e
g
r
a
t
e
[
1
,
{
y
,
0
,
h
}
,
{
x
,

S
q
r
t
[
2
y

2
y
]
,
S
q
r
t
[
2
y

2
y
]
}
,
A
s
s
u
m
p
t
i
o
n
s
0
<
h
<
2
]

(
2

h
)
h
+

(

2
+
h
)
3
h
+
2
A
r
c
S
i
n
h
2
O
u
t
[
]
=
This plot shows how ABC goes from 0 to
π
as
h
goes from 0 to 2.
P
l
o
t
[
a
b
c
,
{
h
,
0
,
2
}
,
F
r
a
m
e
T
r
u
e
,
F
r
a
m
e
L
a
b
e
l
{
"
h
"
,
"
A
B
C
"
}
,
P
l
o
t
L
a
b
e
l
"
A
r
e
a
B
e
n
e
a
t
h
C
h
o
r
d
a
s
a
f
u
n
c
t
i
o
n
o
f
h
"
]
O
u
t
[
]
=
It turns out that Solve can express
h
as a Root.
S
o
l
v
e
[
a
b
c
π
/
4
,
h
,
R
e
a
l
s
]
;
{
%
,
N
@
%
}
/
/
F
l
a
t
t
e
n
h
0
.
5
9
6
…
,
h
0
.
5
9
6
0
2
7
O
u
t
[
]
=
This result passes a basic dummy check.
C
o
l
u
m
n
[
{
π
/
4
,
a
b
c
/
.
F
i
r
s
t
[
%
7
]
}
/
/
N
[
#
,
2
4
]
&
]
0
.
7
8
5
3
9
8
1
6
3
3
9
7
4
4
8
3
0
9
6
1
5
6
6
1
0
.
7
8
5
3
9
8
1
6
3
3
9
7
4
4
8
3
0
9
6
1
5
6
6
1
O
u
t
[
]
=
The point
p
about which the semicircle of cardboard will balance is slightly less than
h
.
E
q
u
a
l
I
n
t
e
g
r
a
t
e
(
p

y
)
2
y

2
y
,
{
y
,
0
,
p
}
,
A
s
s
u
m
p
t
i
o
n
s
0
<
p
<
1
,
I
n
t
e
g
r
a
t
e
(
y

p
)
2
y

2
y
,
{
y
,
p
,
1
}
,
A
s
s
u
m
p
t
i
o
n
s
0
<
p
<
1
;
S
o
l
v
e
[
%
,
p
]
;
{
%
,
N
@
%
}
/
/
F
l
a
t
t
e
n
p

4
+
3
π
3
π
,
p
0
.
5
7
5
5
8
7
O
u
t
[
]
=
A Solution using Region, Area, and RegionCentroid
These results can be checked with RegionCentroid and RegionMeasure.
As expected, RegionCentroid gives the
p
we found above.
{
R
e
g
i
o
n
[
#
]
,
R
e
g
i
o
n
C
e
n
t
r
o
i
d
[
#
]
}
&
@
I
m
p
l
i
c
i
t
R
e
g
i
o
n
[
2
x
+
2
(
y

1
)
<
1
&
&
y
<
1
,
{
x
,
y
}
]
,
0
,

4
+
3
π
3
π
O
u
t
[
]
=
I’ve had no luck finding
h
using Area or RegionMeasure with Solve, NSolve, or FindRoot. It seems that Solve cannot “see inside” ImplicitRegion, though a little voice inside my head says I may be overlooking something
…
S
o
l
v
e
[
R
e
g
i
o
n
M
e
a
s
u
r
e
[
I
m
p
l
i
c
i
t
R
e
g
i
o
n
[
2
x
+
2
(

1
+
y
)
<
1
&
&
y
<
h
,
{
x
,
y
}
]
]
π
4
,
h
,
R
e
a
l
s
]
{
}
O
u
t
[
]
=
The following is a bit clunky, but at least it gives an
h
close to what we found above.
a
r
e
a
I
n
t
e
r
p
o
l
a
t
i
o
n
=
I
n
t
e
r
p
o
l
a
t
i
o
n
[
T
a
b
l
e
[
{
h
,
R
e
g
i
o
n
M
e
a
s
u
r
e
[
I
m
p
l
i
c
i
t
R
e
g
i
o
n
[
2
x
+
2
(
y

1
)
<
1
&
&
y
<
h
,
{
x
,
y
}
]
]
}
,
{
h
,
0
,
1
,
0
.
0
1
}
]
]
;
F
i
n
d
R
o
o
t
[
a
r
e
a
I
n
t
e
r
p
o
l
a
t
i
o
n
[
h
]
π
/
4
,
{
h
,
0
.
5
}
]
;
{
%
,
(
h
/
.
F
i
r
s
t
[
%
7
]
)

(
h
/
.
%
)
}
/
/
F
l
a
t
t
e
n
{
h
0
.
5
9
6
0
2
7
,

1
.
4
6
2
4
5
×

9
1
0
}
O
u
t
[
]
=
A Solution using Mean and Median
Mean and Median give the
h
and
p
found above, and do so with the most concise, and perhaps most transparent, code. Here mass is represented as probability, with mean and median summarizing the PDF as discussed above.
P
r
o
b
a
b
i
l
i
t
y
D
i
s
t
r
i
b
u
t
i
o
n
"
P
D
F
"
,
2
y

2
y
,
{
y
,
0
,
1
}
,
M
e
t
h
o
d
"
N
o
r
m
a
l
i
z
e
"
;
{
M
e
a
n
[
%
]
,
M
e
d
i
a
n
[
%
]
}

4
+
3
π
3
π
,
0
.
5
9
6
…
O
u
t
[
]
=
A Crazy Thought
It’s been said that every problemsolving mistake is either an error, or a blunder [3]. Though most wouldwith some justificationdisagree, I argue that the mistake discussed here is only an error, and a mild one at that. Any solution to any problem has many facets, and being correct (or not) is sometimes the least interesting of a suggested solution's many personality characteristics.
References
[
1
]
h
t
t
p
s
:
/
/
e
n
.
w
i
k
i
p
e
d
i
a
.
o
r
g
/
w
i
k
i
/
C
a
r
_
T
a
l
k
[
2
]
h
t
t
p
s
:
/
/
w
w
w
.
c
a
r
t
a
l
k
.
c
o
m
/
r
a
d
i
o
/
p
u
z
z
l
e
r
/
t
r
u
c
k

b
o
x

a
n
d

m
o
r
e

8
t
h

g
r
a
d
e

m
a
t
h
e
m
a
t
i
c
s

0
[
3
]
H
o
u
s
e
h
o
l
d
e
r
,
A
l
s
t
o
n
S
.
,
P
r
i
n
c
i
p
l
e
s
o
f
N
u
m
e
r
i
c
a
l
A
n
a
l
y
s
i
s
.
D
o
v
e
r
P
u
b
l
i
c
a
t
i
o
n
s
,
1
9
8
1
.
POSTED BY:
John Custy
Answer
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