 # Week 39

Example 1. In the figure given below, the diagonals AC and BD of the quadrilateral ABCD are perpendicular. Given that AB = 8, BC = 7 and DA = 4, find CD.

Example 2. If x and y are any two real numbers, show that (x2 + y2)2xy(x + y)2.

Solution 1. Let O be the point of intersection of the two diagonals AC and BD. Then the four triangles ∆OAB, ∆OBC, ∆OCD and ∆ODA all have right angles at the vertex O. Applying the Pythagorean theorem, we get

OAB  :  (AB)2 = (OA)2 + (OB)2

OBC  :  (BC)2 = (OB)2 + (OC)2

OCD  :  (CD)2 = (OC)2 + (OD)2

ODA  :  (DA)2 = (OD)2 + (OA)2

It follows that

(AB)2 + (CD)2

= (OA)2 + (OB)2 + (OC)2 + (OD)2

= (OB)2 + (OC)2 + (OA)2 + (OD)2

= (BC)2 + (DA)2.

Since AB = 8, BC = 7 and DA = 4, we conclude that

(CD)2 = (BC)2 + (DA)2 − (AB)2

= 72 + 42 − 82

= 1.

Therefore, CD = 1.

Solution 2. We prove the inequality by showing that the left side minus the right side is always

greater than or equal to zero. We compute

(x2 + y2)2xy(x + y)2

= x4 + 2x2y2 + y4x3y – 2x2y2xy3

= x3(xy) – y3(xy)

= (xy)(x3y3).

Since x and y are real numbers, either x = y or x > y or x < y.

If x = y, then (xy)(x3y3) = 0.

If x > y, then x3 > y3. Therefore, (xy)(x3y3) > 0.

If x < y, then x3 < y3. Therefore, (xy)(x3y3) > 0.

Thus, in any case, (xy)(x3y3) ≥ 0.

It follows that, (x2 + y2)2xy(x + y)2.