Download Module 5 Class Notes

MATH4455
Module 5: Tri, Tri Again
Main Math Concepts: Triangles, Laws of Sines and Cosines, Congruence and Similarity, Medians
Auxiliary Ideas: Classification, Parameter diagrams
I. Specifying Triangles.
Generally, we can specify ABC up to congruence by giving three independent scalar values.
Problem a. (SSS) Two sides of a triangle are a = 6 and b = 8. The third side is c.
What values of c are possible? Classify the triangles so obtained as Right, Obtuse, Acute. For which
values of c is the triangle isosceles? What happens at non-isosceles-values of c ?
Solution. Since the sum of any two sides must exceed the other side, then 2 < c < 14. There are
degenerate triangles at c = 2 and c = 14.
For c < 8, a right triangle will happen when c 2 + 6 2 = 8 2 , i.e. when c = 28 = 2 7 ≈ 5.3 . For c smaller
than that, the triangle is obtuse since then c 2 + 6 2 < 8 2 . As c grows larger than 2 7 , the triangle is acute
until c reaches 6 2 + 8 2 = 10 , where c is the hypotenuse of another right triangle. Beyond that, the
triangles are obtuse until the triangle degenerates to a line at c = 14.
We also note that the triangle is isosceles at c = 6 and at c = 8. These values mark off the regions where
the ordering of the three sides changes. A sketch summarizing all this:
(sides order)
| c<a<b
I a<c<b I
a<b<c
|
(triangle none | obtuse R
acute
R obtuse
| none
type) _____________________________________________________
|
5.3
|
c
0
2
4
6
8
10
12
14
I denotes Isosceles at c =
6, 8. R denotes Right
triangle at c=5.3, 10.
There are no triangles
for c ≤ 2 or c ≥ 14.
Problem b. (ASA) Suppose that angle C is 60 degrees, and side b has length 6.
i) Which angles A are possible, and classify the triangles into Right, acute, obtuse, Isosceles, etc.
ii) Find side a and the area in terms of angle A.
Solution.
Since the sum of the angles is 180, we must have 0 < A < 120 . We get Right triangles when A = 30 or
A = 90, acute when 30 < A < 90, and obtuse otherwise. For an isosceles triangle, we need that two angles
be equal which amounts to either B = A = 60 or B = C = (180 – 60)/2 = 60, that is, the only time we get
an isosceles triangle here is when B = 60, in which case the triangle is equilateral. Summary diagram is
on the next page.
|
a<c<b
I
b<c<a
|
c
a
60
triangle || obtuse
R
acute
R obtuse | none
_____________________________________________________
|
|
|
|
A
0
30
60
90
120 (degrees)
6
A
ii) We could drop an altitude, but we would be deriving the Law of Sines, which we’ll just use directly:
a
6
c
a
b
c
=
=
, so we have
and then
=
=
sin A sin(120 − A) sin 60
sin A sin B sin C
a=
6 sin A
6 sin A
6 sin A
=
=
=
sin(120 − A) sin120 cos A − sin A cos120 ( 3 / 2)cos A + (1 / 2)sin A
12 sin A
3 cos A + sin A
and
a
.
h
60
1
3 3 / 2(12 sin A)
18 3 sin A
(6)(h) = 3(a sin 60) =
=
.
2
3 cos A + sin A
3 cos A + sin A
------------------------------------------------------------------------AREA =
b=6
A
Summary of specifying triangles, up to congruence, by three values from the sides or angles.
Name Given data
# triangles up to congruence
SSS
a, b, c > 0
1 if sum of any two exceeds the third, otherwise 0.
SAS
a, b > 0; 0 < C< 180 1 always
ASA
0 < A, B <180; c > 0 1 if A + B <180, otherwise 0
ASS
0 <C<180; b ,c > 0
If C ≥ 90, then: 1 if c > b, 0 otherwise
⎧
⎨
⎩If C < 90, then 0, 1, or 2 depending on b, c, and C
The subcase C < 90 for ASS case is homework problem #1
Note that since the sum of the angles in a triangle is always 180, there are really only two angles
available to be specified in a triangle, since once you know two you automatically know the third. So the
four cases given above are all the ways to give three independent scalars for a triangle (e.g. AAS and
SAA are the same as ASA since you can find the third angle, and SSA is the same as ASS by reordering).
II. The Ratio Plane
Many interesting properties of triangles depend on their shape rather than their size, and we will focus on
triangle shapes for the rest of this class. By the shape of a triangle, we mean that ABC and  A′B′C ′
have the same shape if they are similar, which can be characterized in many ways, the most common
a ′ b ′ c′
being (i)
and (ii) A = A′, B = B', C = C ′ . Of course, we only need to verify two of the
= =
a b c
three angles in (ii), since then the third follows since the sum of the three is 180 in both triangles.
Specifying a triangle shape generally requires two scalar values—the obvious one being two angles. For
example, there is one triangle shape with A = 30 and B = 60 (namely the 30-60-90 shape). However
there are other ways to specify a triangle shape with two scalars.
For example, the shape of the triangle with sides (a, b, c ) = ( 6, 8, 12 ) could be defined by the ratios x =
a / c = 1/2 and y = b / c = 2/3. that is, the triangles ABC which are similar to the one with sides (6, 8, 12)
are precisely the ones with x = 1/2 and y = 2/3, where x = a / c and y = b / c. This is easy to check since
a b c
a 6 1
if ABC is similar to the given triangle, then = =
and then solving gives =
= and similarly
6 8 12
c 12 2
b 8 2
=
= , and the converse is equally straightforward. Note that any unlabeled scalene shape has six
c 12 3
different labelings so there are six different pairs (x,y) corresponding to that one unlabeled shape.
Problem a. Define the following sets of triangle shapes in terms of the ratios x=a/c and y=b/c. By
convention, the order of the given sides is (a,b,c) and the order of given angles is (A,B,C) unless specified
otherwise.
i) The shape of the triangle (6, 8, 12)
ii) The shape of a 30-90-60 triangle.
iii) The shape of a 30-60-90 triangle
iv) The set of shapes of triangles (6, 8, c )
Solution. (i) as above, this single shape is defined by (x, y ) = (1/2 , 2/3).
ii) The sides of a 30-90-60 triangle are (multiples of) (1, 2,
3 ), so the ratios are x =
1
2
and y =
, so
3
3
(x, y) = (1 / 3, 2 / 3) .
iii) This shape is like (ii) except that the order of the vertices is different, so we have the sides (a,b,c)
beings multiples of (1, 3 , 2), so we get (x, y) = (1 / 2, 3 / 2) .
iv) This is a set of shapes, not just one shape, and we have x =6 / c, y =8 / c which amounts to saying that
y / x = 8 / 6 or simply y = (4/3) x.
.
Problem b. In the x-y plane, sketch the region of points (x,y) which are triangle ratios x = a/c, y = b/c,
and inside that region, determine which points correspond to isosceles shapes and the various orderings
of the side lengths. Finally plot the points corresponding to shapes (i)-(iv) in Problem (a) above.
Solution. Given (a,b,c), they form a triangle if the sum of any two exceeds the third. We can divide
through by c to get three inequalities in x and y:
a+b>c
a+c>b
b+c>a
⇒ a / c + b / c > 1 ⇒ x + y > 1, i.e. y > 1 − x
⇒ a / c + 1 > b / c ⇒ x + 1 > y, i.e. y < 1 + x
⇒ b / c + 1 > a / c ⇒ y + 1 > x i.e. y > x − 1
So the region for triangle ratios lies above the lines y = 1 – x and y = x – 1, and below the line y = x + 1.
For isosceles triangles, we have three different possibilities:
a = c, which is equivalent to x = 1
b = c, which is equivalent to y = 1
a = b, which is equivalent to a/c = b/c, i.e. x = y.
So the isosceles shapes correspond to the lines x = 1, y = 1, and x = y.
Here are the graphs, with the points from Prob. IIa included on the right. The line on the right is y=(4/3)x
which corresponds to the (6,8,c) shapes, and the three points are ( 1/2 , 2/3), (1 / 3, 2 / 3) , and
(1 / 2, 3 / 2) .
y=b/c
y=b/c
4
2.0
3
1.5
2
1.0
1
0.5
1
0.5
1.0
1.5
Outer lines enclose the region
of triangle ratio points. The
three inner lines are the ratio
points for isosceles triangles.
2.0
2
3
x = a/c
The three points and the line
corresponding to the examples in
Problem IIa, (along with the
boundary and isosceles lines) .
4
x= a/c
III. The Family of 60 Triangles.
Problem a. Consider the set of shapes in which angle C is 60 .
(i) Suppose c = 7 and b = 8. What is the third side a ?
(ii) ). Find the maximum possible value of y = b/c in a triangle where C = 60 ? What particular shape is
that?
(iii) Find and sketch the set of points in the x-y ratio plane corresponding to the family of triangle
(shapes) in which C = 60 . Indicate the points corresponding to triangles in (i) and (ii).
Solution. We need a relationship between the sides of the triangle knowing that angle C = 60 . The tool
for that task is the Law of Cosines: c 2 = a 2 + b 2 − 2ab cosC . Since cos 60 = ½, we have in this case
c 2 = a 2 + b 2 − ab .
(i) Given that c = 7 and b = 8, we have 49 = a 2 + 64 − 8a, so a 2 − 8a + 15 = 0 which factors into
(a – 3)(a – 5) = 0, so a= 3 or a = 5. Thus, there are two such triangles with sides (3, 8, 7) and ( 5, 8, 7).
The shape ratios of these triangles are (3/7, 8/7) and (5/7, 8/7).
(ii) If we imagine side b as fixed, then to maximize y = b / c amounts to minimizing c and you can see
from the picture on the right, that the minimum c occurs when angle B is a right angle. Just look at the
triangle and think about the minimum possible c for a given b. Then we have a 30-90-60 triangle and b/c
is (hypotenuse/ (long leg)) = 1 / sin(60 ) = 2 / 3 .
a 2 b 2 ab
2
iii) Dividing the equation above by c gives 1 = 2 + 2 − 2 = x 2 + y 2 − xy , that is, x 2 + y 2 − xy = 1 .
c
c
c
This equation is similar to Prob. #4 in HW 2. It is symmetric about the line y= x, has positive intercepts
(1,0) and (0,1) with maximum y = b/c = 2 / 3 which occurs at x = a / c = 1 / 3 . which is just the 3090-60 shape from (ii). The graph is x 2 + y 2 − xy = 1 restricted to the region of triangle ratios.
y
1.4
1.2
(a =3, 4, 5, 8)
3
c=8
1.0
1
0.8
c=7
1
0.6
a=3
0.4
c=7
60
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
x
C
b=8
A
The triangles on the right have the four shapes with C = 60 shapes marked on the graph, namely:
(1,1) [i.e.equilateral]; (5/7, 8/7),;
(3/7, 8/7)
(1 / 3, 2 / 3) [i,e, 30-90-60];
Looking at the graph above, we can see that except for the right triangle 30-90-60, the other shapes with
a < c < b come in pairs which have the same y= b/c ratio and two different x = a/c values. We refer to
these two shapes as 60 twins, e.g. (3,7,8) and (5,7,8) are “ 60 twins” in that they have the same angle
C= 60 and the same y ratio 8/7.
Problem b. Find the 60 twins with b = 15, c = 13, and with b = 10, c =9. What seems to be true about a
and a′ in the C = 60 twin triangles (a,b, c) and (a′,b, c) ? Give both a geometric and an algebraic
explanation of this relationship.
Solution. For triangles with C = 60 we have c 2 = a 2 + b 2 − ab, or a 2 − ab + b 2 − c 2 = 0 .
b = 15, c = 13: a 2 − 15a + 225 − 169 = 0 = a 2 − 15a + 56 = (a − 8 _(a − 7) . So a = 7 or a = 8.
b = 10, c = 9: a 2 − 10a + 100 − 81 = 0 = a 2 − 10a + 19; a =
10 ± 100 − 4(19)
= 5 ± 6 . So a = 5 ± 6
2
In all three of our twin examples, the sum of the two values of a equals b, more precisely:
Conjecture: If (a,b, c) and (a′,b, c) are twin triangles with C= 60 , then a + a′ = b .
Algebraic proof: By the Law of Cosines we have c 2 = a 2 + b 2 − ab, or a 2 − ab + b 2 − c 2 = 0 .
By quadratic formula, the two values of a are a =
b ± b 2 − 4(b 2 − c 2 ) b
4c 2 − 3b 2
= ±
.
2
2
2
b
4c 2 − 3b 2
b
4c 2 − 3b 2
b b
The sum of the two values of a are a + a′ = ( +
)+( −
)= + =b
2
2
2
2
2 2
(or you could remember that the sum of the roots of a monic quadratic equals the negative of the middle
term).
Geometric proof. The two twin triangles can be constructed by drawing
an arc of length c from vertex A. Also draw the perpendicular from A to
the opposite side intersecting BC at point D. The right triangle ACD is a
30-60-90 so CD = b/2. Furthermore, the lengths B’D and BD are equal
as both are z =
B
c 2 − AD 2 . Then
QED.
Notice that z is just the term z =
B’
4c − 3b
from the algebraic proof.
2
2
2
Also, we can see from the picture that we won’t get these twins if c > b
– in that case, one of the two values of a would be negative, and the
twins would be of the form (a, b, c) and ( a, b ‘, c ) since then b < c < a.
z
D
a′ = b / 2 − z , a = b / 2 + z , so a′ + a = b / 2 − z + b / 2 + z = b.
c
z
c
C
b
A
IV. Medians and Another Family
A median of a triangle is the line from a vertex to the midpoint of the opposite side, so any triangle has
three medians, one emanating from each vertex.
Let D be the midpoint of side AC, let E be the midpoint of BC. and let F be the intersection of the
medians AE and BD.
Problem a. Find two pairs of similar triangles in this picture for each pair, find the ratio of the
corresponding sides. If u is the length of AF, find AE in terms of u.
Solution. It looks like CDE is similar to CAB, and indeed we have CA = 2 CD, CB = 2 CE, and then by
the Law of Cosines we have
a
b
ab
a
b
ab
c 2 = a 2 + b 2 − 2ab cosC = (2 )2 + (2 )2 − 2(4)( )cosC = 4[( )2 + ( )2 − 2( )cosC] = 4(DE)2
2
2
22
2
2
22
2
2
Thus, c = 4(DE) , so c = 2(DE) and thus we have that CDE is similar to CAB by a factor of ½.
Hence, CDE = CAB and CED = CBA , and so DE is parallel to BA by corresponding angles.
Now, with DE parallel to BA, we have that FDE = FBA and FED = FBA by alternate interior
angles. It follows that triangle FED is similar to FAB. The factor must again be ½, since we already
know that (DE ) = 0.5 ( AB) and those are corresponding sides of the present pair of triangles.
C
Finally, EF and AF are corresponding
sides of similar triangles with factor ½,
a/2
then we know that EF = 0.5 u, and so the
b/2
length of the median AE = 1.5 u.
Equivalently, AF = (2/3) AE.
D
E
b/2
A
u
F
c
By the same computation, FD = 0.5(BF)
and BF = (2/3) BD. So we have derived:
a/2
B
Any two medians of a triangle intersect
at a point which is 2/3 of the way from
each vertex to the opposite side.
It also follows that all three medians of a triangle intersect at the same point, since, e.g. the median from
C must intersect the median from A at a point 2/3 of the way from A to E by what we just showed. But
that is also the point where the median from B intersects the median from A. Thus they all meet at a
single point, 2/3 of the way from each vertex to the other end of the median, called the centroid of the
triangle.
In general the triangles, AFD and BFE are not similar unless triangle ABC is isosceles.
We now define an interesting family of shapes using the medians. Say that a triangle ABC is median
perpendicular at C (MPC) if the medians from vertices A and B are perpendicular to each other. The
property MPC is a “shape” property in that if two triangles are similar, then they are both MPC or both
not MPC.
Problem b. Characterize the family of MPC triangles in terms of the sides a, b, and c, and use it to graph
the (x,y) points corresponding to the ratios for MPC triangles.
Solution. Suppose that ABC is an MPC triangle. Label D, E , F and u as we did in the last example, and
let v be the length of FB. Then, since we are working with medians, we have
FE = u / 2 , FD = v / 2, DE = c / 2, AD = b / 2, BE = a / 2 .
However, in the present case we also have the all four angles at F are right angles, so we get
(i) c 2 = u 2 + v 2
b
v
(ii) ( )2 = ( )2 + u 2 from right triangles at F. We want to eliminate u and v.
2
2
a
u
(iii) ( )2 = ( )2 + v 2
2
2
a2
u 2 3u 2
= u2 −
=
, so we have 4c 2 − a 2 = 3u 2
Subtracting (iii) from (i) gives c 2 −
4
4
4
Multiplying (ii) by 4, then subtracting (i) from that gives b 2 − c 2 = v 2 + 4u 2 − (u 2 + v 2 ) ⇒ b 2 − c 2 = 3u 2
Equating the expressions for 3u 2 gives us 4c 2 − a 2 = b 2 − c 2 ⇒ 5c 2 = a 2 + b 2
a2 + b2
, i.e. 5c 2 = a 2 + b 2 .
So we have that the MPC triangles are the ones for which c =
5
2
a
b2
We can convert this equation to shape ratios by dividing: 5 = 2 + 2 = x 2 + y 2 and so the set of ratio
c
c
points corresponding to MPC triangles is the circle of radius 5 about the origin.
C
2
2.0
b/2
a/2
D
b/2
The C=60 degree triangle
curve is also shown. We can
see that the angle C in an
MPC triangle must be (much)
smaller than 60 degrees, as
the MPC curve is so much
further away from (0, 0).
Notice that all MPC triangles
have c as the shortest side (by
the regions in which the curve
lies).
1.5
E
v/2
u/2
a/2
u F v
1.0
0.5
A
c
B
0.0
0.0
0.5
1.0
1.5
2.0