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)2 – e.
Clearly with this pattern, the next row is 2 × 292 = 412 + 1.
Now, we need to show that 2(a + b)2 = (2a + b)2 – e 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)2 – e]
= 2a2 + 4ab + 2b2 – (4a2 + 4ab + b2 – e)
= b2 – 2a2 + e
= 0.
It follows that 2(a + b)2 = (2a + b)2 – e, 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°. [4]
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.
Math1089 