For in which sides
have slopes
, respectively, the equation connecting their side lengths is
(1)
as we saw a post ago. The equation connecting their median lengths is
(2)
and is today’s goal.
Compare the relationship between the medians of a right triangle with the medians of a triangle with slopes .
This is the final, knockout round of our compare-contrast-contest with the right triangle. The contest started on June 28, with the right triangle currently having an edge.
- In a right triangle, the medians are related via
, where
is the length of the median to the hypotenuse (see the exercises). The triangle with slopes
doesn’t satisfy this relation, unless we put
in (2).
Winner: Right triangle; Score is.
- The previous score was
, in favour of the right triangle.
Final verdict: Right triangle wins on aggregate. Case closed.
We still celebrate the triangle with slopes , despite been rightly trounced by the right triangle.
Consider with vertices
,
, and
. Verify that three medians satisfy equation (2).
Note that the slopes of sides are
, respectively. They form a geometric progression with
. The midpoints of these sides have been marked. Thus, by the distance formula:
YEA.
Consider with vertices at
,
, and
. Verify that the three medians satisfy equation (2).
Easy-peasy. Observe that the slopes of sides are
. They form a geometric progression with common ratio
. Let’s draw a diagram:
The median lengths then follow by the distance formula. We have:
There we go. The formula works for both positive and negative common ratios.
In , the median lengths are related via
. Find (possible) coordinates for the vertices
.
Try solving this problem in a way other than what we present below.
In (2), set and solve for
:
The workload reduces to finding coordinates for a triangle whose slopes form a geometric progression with common ratio .
Easy-peasy.
Among the many options we have for the vertices of a triangle with side slopes , we can take
,
, and
With
, these become
,
, and
. Use these as
. That is, take
,
, and
as the triangle’s vertices. A quick calculation shows that
,
, and
(as in, triple three).
Watch what happens when in both (1) and (2).
Put in (1). Obtain
.
Put in (2). Obtain
.
These are well-known, equivalent conditions for the medians from and
to be perpendicular.
Is it surprising that it’s our favourite -value that saved the day?
For , let
. Evaluate
, and show that they are all sums of two consecutive squares.
Observe that the only integer values of that produce integer values of
are
.
In addition, defines a bijection
.
In , we have
and
. Find the value of
.
You can suspect what will be. From (1) and (2), set
Clear fractions, and simplify to obtain the quartic polynomial equation
The second quadratic factor, , is really irreducible (its discriminant is
). So the only real roots come from the first one:
Thus, for , we have:
Similarly, for , we have:
The value of is
, as you may have correctly guessed.
OK, the above approach was definitely an overill for something that can be obtained in a simpler way, but it just confirms that our theory conforms to what is more established. See the next example for a better strategy.
In any , PROVE that the following two statements are equivalent:
- the relations
and
hold simultaneously;
.
Consider and let’s default to standard notation. Use the fact that the median lengths are related to the side lengths via:
(3)
from which . Now, since
, this becomes
. Also,
becomes
. Since
, we have:
In view of , we see that
can’t be negative (in fact,
: example 10), so discard
, and take
as the only valid solution. Thus,
.
To see that , suppose that
. Assume that
. We’ll show that
as well. Indeed, from (3):
Similarly, if we started with , we’ll again arrive at
. This proves
.
Bottom line: the two statements are equivalent as claimed.
Kindly bear with us for switching back and forth between the standard notation and
. Hopefully this doesn’t cause any confusion.
In any , if
and
, PROVE that
.
The relationship
holds in any triangle. Using the fact that and
, we have:
Although we freely cancelled out the common factor , it shouldn’t obscure the fact that the relations
and
hold simultaneously only when
.
If , PROVE that
.
Compare with exercise 5 below.
For the proof, begin with one of the expressions in (3):
Since the left side is positive, so must the right side be, and we only have to impose this restriction on the nonkonstant
.
, as desired.
Takeaway
In the equation
the median from vertex and the median from vertex
are perpendicular only when
. If the focus shifts to the medians from
and
, or the medians from
and
, then other values of
can still ensure perpendicularity.
Tasks
- Find possible coordinates for the vertices of
if its three medians are related via
.
- Find possible coordinates for the vertices of
if its three medians are related via
.
- PROVE that the medians in a right triangle are related via
, where
is the length of the median to the hypotenuse.
- Derive equation (2).
- Let
and
. PROVE that both
and
have the same minimum value, but that their common minimum occurs at different values of
.
- Consider the rational function
:
- Verify equation (2) for a triangle with vertices at
,
, and
.
- PROVE that if a right triangle has slopes in geometric progression with common ratio
, then both the squares of the median lengths and the squares of the side lengths form arithmetic progressions.
- PROVE that the “mapping”
, is not well-defined. What modifications should be made to the domain or range, so that the mapping can be well-defined?
- If
and
, PROVE that
is then a sum of two consecutive squares.