# Week 42

Example 1. Observe that

2 × 12          = 12 + 1

2 × 22          = 32 – 1

2 × 52          = 72 + 1

2 × 122        = 172 – 1.

Show that there are actually many positive integer solutions to each of the equations 2x2 = y2 + 1 and 2x2 = y2 – 1.

Example 2. Altitudes AP, BQ and CR are drawn in ∆ABC, and these lines meet at point H, as indicated. Suppose that AH = BC. Show that PR and PQ are perpendicular.

Solution 1. First, we need to find the pattern here.

If we write any row as 2a2 = b2 + e, then we have either e = 1 or e = –1.

The left side of the next row looks like 2(a + b)2 and the right side of the same is (2a + b)2e.

Clearly with this pattern, the next row is 2 × 292 = 412 + 1.

Now, we need to show that 2(a + b)2 = (2a + b)2e whenever 2a2 = b2 + e.

This will show that the pattern continues indefinitely, and hence there are infinitely many solutions to each of the two equations 2x2 = y2 + 1 and 2x2 = y2 – 1.

Assume that 2a2 = b2 + e. Then

2(a + b)2 – [(2a + b)2e]

= 2a2 + 4ab + 2b2 – (4a2 + 4ab + b2e)

= b2 – 2a2 + e

= 0.

It follows that 2(a + b)2 = (2a + b)2e, as expected.

Solution 2. Since AP is an altitude, ∆APC is right-angled with ∠APC = 90°.

Therefore, ∠HAQ + ∠ACB = 90°.                           (1)

Similarly, working in the right ∆BQC, we see that ∠CBQ + ∠ACB = 90°.                    (2)

It follows that ∠HAQ = ∠CBQ.                  (3)

In ∆AHQ and ∆BCQ, we have

AQH = 90° = ∠BQC

AH = BC

HAQ = ∠CBQ  [by (3)]

Therefore, ∆AHQ ≅ ∆BCQ  [by SAA].

Thus BQ = AQ, so ∆AQB is an isosceles right triangle.

Hence, ∠ABQ = ∠RBH = 45°.                     

Now consider the four points B, R, H and P.

Since ∠BRH = 90°, it follows that point R lies on the unique circle that has BH as a diameter. Following the same argument, point P also lies on this circle.

Thus, since ∠RBH and ∠RPH are inscribed in this circle and subtend the same arc, it follows that ∠RPH = ∠RBH = 45°  [by (4)].

Exactly similar reasoning shows that ∠QPH = 45°, and thus ∠RPQ = 90°, as required.