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Suggested problems - solutions
Taxicab geometry
Material for this section references College Geometry: A Discovery Approach, 2/e, David C. Kay, Addison Wesley, 2001. In particular, see section 3.2, pp 127-137.
The problems are all from section 3.2.
Problems: 1, 2, 3, 4, 5, 6, 10, 11
#1, #2 Distance in the Taxicab metric is given by
P Q∗ = |x2 − x1| + |y2 − y1|
- the sum of horizontal and vertical distance.
4ABC:
AB ∗ = 0 + 6 = 6
BC ∗ = 6 + 0 = 6
CA∗ = 6 + 6 = 12
4XY Z:
XY ∗ = |8 − 5| + |8 − 5| = 3 + 3 = 6
Y Z ∗ = |5 − 8| + |11 − 8| = 3 + 3 = 6
ZX ∗ = |5 − 5| + |5 − 11| = 0 + 6 = 6
It is true that AB ∗ = XY ∗ and BC ∗ = Y Z ∗ , and the angle included between the sides is a right
angle in both triangles, so 6 B ∼
= 6 Y and SAS = SAS (the SAS hypothesis). However, we don’t
get the familiar Euclidean SAS conclusion: 4ABC is NOT congruent to 4XY Z, as this would
mean that all the remaining corresponding sides and angles were equal as well. And the third
sides are not equal: CA∗ = 12 and ZX ∗ = 6.
Note: don’t go by looks to say the triangles aren’t congruent - if we’re looking at congruence in
Taxicab geometry, we want corresponding sides to have equal measures in Taxicab geometry,
and we know that visual “distance” and Taxicab “distance” don’t always match up.
Case in point: 4XY Z is equilateral under the Taxicab metric (it cannot, however, be equiangular, so there’s another Euclidean theorem gone).
And, under the Taxicab metric, 4ABC is isosceles, with base CA.
#3 Given an equilateral triangle:
You can verify this by calculating the side lengths in the Euclidean metric using the Pythagorean
theorem/distance formula- the length of 2 marked on the hypotenuse is the Euclidean length,
not the Taxicab length.
(a) AB ∗ = 2, BC ∗ = |1 − 0| + |0 −
√
√
√
√
3| = 1 + 3, CA∗ = |0 − 1| + | 3 − 0| = 1 + 3
(b) The angle measures have nothing to do with the Taxicab values; they come from Euclidean
measure. Since the triangle is equilateral in Euclidean measure, it is equiangular, and the
angles measure 60◦ .
(c) In Taxicab geometry, 4ABC is isosceles, with base AB, since BC ∗ = CA∗. It is also
equiangular in Taxicab geometry, since once the angle values are established using Euclidean
measure, they become the Taxicab values as well.
#4 The measures of the sides in the Taxicab metric are marked on the diagram:
(a) Since RS ∗ = ST ∗ = T R∗ = 16, 4RST is equilateral.
(b) You can tell from inspection it isn’t, but to do it right, you need to calulate the Eucilidean
lengths:
√
√
52 + 112 = 146
RS =
√
√
ST =
82 + 82 = 128
√
√
TR =
132 + 32 = 178
Since the triangle is scalene under Euclidean measure, none of the angles are equal. So none
of the angles are equal in Taxicab measure. (So yes, we can have an equilateral triangle
with no equal angles in this geometry)(.
(c) Did the side lengths above - triangle is scalene under the Euclidean metric.
#5 The measures of the sides in the Taxicab metric are marked on the diagram:
9 − 0 = −3, and m = 4 − 0 = 1 . Since −3( 1 ) = −1 (the slopes are negative
(a) mAC = −3
BC
−0
12 − 0
3
3
reciprocals), AC ⊥ BC and 4ABC is a right triangle, with right angle 6 C.
(b) On diagram (I think it’s easier to do the calculations visually, instead of plugging into the
formula - go with whatever works for you).
c = AB ∗ = 20
b = AC ∗ = 12
a = BC ∗ = 16
(c) Does a2 + b2 = c2?
a2 + b2 = 122 + 162 = 144 + 256 = 400 = 202
Yes. Pythagorean relation holds.
#6 The measures of the sides in the Taxicab metric are marked on the diagram:
OK, this one’s obvious - no, 72 6= 32 + 42 . No Pythagorean relation for this triangle (so, once
again, another theorem failing to work in general...)
#10 I’d use a combination of algebra and geometry on this one - the relationships from algebra are
|x − 1| + |y − 0| = |x − 7| + |y − 2| = |x − 8| + |y − 3|
which can be used to produce pairs of equations:
|x − 1| + |y| = |x − 7| + |y − 2| and |x − 1| + |y| = |x − 8| + |y − 3|
The geometry comes in in that I don’t want to consider every possible case for those absolute
values! Taking the marked R in the text as an estimate (to the right of W , left of L and E, and
above all three points), the first equation becomes:
x − 1 + y = 7 − x + y − 2 → x − 1 = 5 − x → 2x = 6 → x = 3
and then
|3 − 1| + y = |3 − 8| + y − 3... hey, I can’t solve that!
Ah, those sneaky misleading diagrams...
How about we try it right of W , left of L and E, above W and L, below E?
x−1+y = 7−x+y−2
x−1+y = 8−x+3−y
→
2x + 2y = 12
→
2x = 6
→
→
x+y = 6
x=3
→
3+y = 6
→
y=3
Let’s see how (3, 3) works ...
There - a 5 block walk to any of the places. (It turns out to be level with E, but I needed the
initial assumption to make the absolute values work out.)
You could also have done this visually with some trial and error, which is fine, too. I like looking
at the algebraic approach, though.
#11 This one is also an algebra or a geometry problem - take your pick. To solve geometrically, make
an accurate sketch, sit down with a compass, and construct some perpendicular bisectors:
I get the intersection of the perpendicular bisectors located at about (2, 8) ... ish. (Haven’t
figured out yet how to make annoying construction software give point of intersection ... but if
you’re doing pencil and paper, you’re doing it visually too ...)
Doing this algebraically - get midpoints:
midW L =
midLE =
Get slopes:
1+7 0+2
,
= (4, 1)
2
2
7+8 2+3
,
= (7.5, 2.5)
2
2
2−0
1
=
7−1
3
3−2
=1
mLE =
8−7
mW L =
Get line through (4, 1) with slope -3 (⊥ to W L):
y − 1 = −3(x − 4)
→
y − 1 = −3x + 12
→
y = −3x + 13
Get line through (7.5, 2.5) with slope -1 (⊥ to LE):
y − 2.5 = −1(x − 7.5)
→
y − 2.5 = −x + 7.5
→
y = −x + 10
And intersect lines:
−3x + 13 = −x + 10
→
y = −1.5 + 10
−2x = −3
→
→
x = 1.5
y = 8.5
Intersection is (1.5, 8.5). That’s the rough bit - you can calculate how many blocks from there.