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?
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.
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.
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.
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.
Let CA = a and CB = 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
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.
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.
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.
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.
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.
The gradient of BB' = s/(r - 2).
The gradient of AA' = (s/2) / (r/2 - 4) = s/(r - 8).
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.
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 .
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