Solution to puzzle 100: Perpendicular medians

Suppose the medians AA' and BB' of triangle ABC intersect at right angles. If BC = 3 and AC = 4, what is the length of side AB?

A number of approaches to this problem are possible.

Geometric Solution

In the diagram below, BCA = A'CB', and CA'/CB = CB'/CA = ½.
Hence triangles CAB and CB'A' are similar; and A'B' = ½BA.
Let AA' and BB' intersect at D.
Let A'D = x, B'D = y, AD = z, BD = w. Let AB = c, so that A'B' = ½c.

Triangle ABC, with perpendicular medians AA' and BB', intersecting at D, and line segment A'B'. Length AB=c, A'B'=c/2, A'D=x, B'D=y, AD=z, BD=w.

Applying Pythagoras' Theorem to each of the four right-angled triangles shown in the diagram:

A'B'D	y² + x² = c²/4.	(1)
B'AD	y² + z² = 4.	(2)
ABD	w² + z² = c².	(3)
BA'D	w² + x² = 9/4.	(4)

Then (1) − (2) + (3) − (4) 0 = 5c²/4 − 25/4.
Hence c² = 5.

Therefore the length of side AB is .

Vector Solution

Let C be the origin.
Let CA = a and CB = b.

Triangle ABC, with perpendicular medians AA' and BB'. CA = vector a; CB = vector b.

We now determine vectors AA' and BB', and, as they are perpendicular, set their dot product equal to zero.

We have AA' = AC + CA' = ½b − a.
Similarly BB' = BC + CB' = ½a − b

AA' BB' AA' . BB' = 0.
Hence (b − 2a).(a − 2b) = 0.
And so 5a.b − 2a² − 2b² = 0.
Since a = 4 and b = 3, we obtain 5a.b = 2(4² + 3²) = 50.
Hence a.b = 10.

Now, AB . AB	= (b − a).(b − a)
	= b² + a² − 2a.b
	= 3² + 4² − 2×10
	= 5.

Therefore the length of side AB is .

Geometric Solution using a Property of Medians

Let AA' and BB' intersect at D.
In this solution we use the well known result that the medians of a triangle intersect 2/3 of the way from the vertex to the midpoint of the opposite side.
We may therefore let A'D = x, DA = 2x; and B'D = y, DB = 2y.

Triangle ABC, with perpendicular medians AA' and BB', intersecting at D.

Applying Pythagoras' Theorem to each of the three right-angled triangles shown in the diagram:

A'DB	x² + 4y² = 9/4.	(1)
ADB'	4x² + y² = 4.	(2)
ABD	4x² + 4y² = c².	(3)

Then (1) + (2) 5x² + 5y² = 25/4.
Substituting into (3), we obtain c² = (4/5) × (25/4) = 5.

Therefore the length of side AB is .

Cartesian Solution

Less elegant (though quite short!) is a Cartesian solution. We use the fact that the product of the gradients of perpendicular lines is equal to −1.

Let C be the origin. Let the coordinates of A be (4, 0), and those of B be (r, s). Note that the gradients of line segments AA' and BB' must have opposite signs, and hence r > 2.

Triangle ABC, with perpendicular medians AA' and BB'. B = (r,s), A = (4,0), B' = (2,0), A' = (r/2,s/2).

The gradient of BB' = s/(r − 2).
The gradient of AA' = (s/2) / (r/2 − 4) = s/(r − 8).

AA' BB' the product of the gradients of AA' and BB' equals −1.
Hence s²/(r − 2)(r − 8) = −1.
Simplifying, we obtain s² = −r² + 10r − 16.
Since BC = 3, we also have r² + s² = 9, so that 10r − 16 = 9, and r = 5/2.

Then AB²	= s² + (4 − r)²
	= s² + r² − 8r + 16
	= 9 − 20 + 16
	= 5.

Therefore the length of side AB is .

Source: A Survey of Classical and Modern Geometries (Exercise 1.69), by Arthur Baragar. With thanks to Paul M. Hether for bringing this puzzle to my attention, and to Michael Hemy for suggesting the first (and best) solution, above.