Let ABCD be a rectangle such that AB=5 and BC=12. There exist two distinct points X1 and X2 on BC such that ∠AX1D=∠AX2D=90∘. Suppose that d is the distance from X1 to X2. What is d2?

Question

Let ABCD be a rectangle such that AB=5 and BC=12. There exist two distinct points X1 and X2 on BC such that ∠AX1D=∠AX2D=90∘. Suppose that d is the distance from X1 to X2. What is d2?

Answer 1

To find the value of d^2, we need to determine the distance between points X1 and X2.

Since ∠AX1D = ∠AX2D = 90°, it implies that X1D and X2D are both perpendicular to AD. In a rectangle, diagonals are equal in length and bisect each other.

So, AD is the diagonal of the rectangle and it divides BC into two equal parts. Let's label the point where AD intersects BC as E.

Now, we have two right-angled triangles, namely ΔAX1D and ΔAX2D.

In triangle ΔAX1D, AD is the hypotenuse, and since it is equal to the diagonal of the rectangle, its length is the square root of the sum of the squares of the rectangle's sides:
AD = sqrt(AB^2 + BC^2)
= sqrt(5^2 + 12^2)
= sqrt(169)
= 13

Similarly, in triangle ΔAX1D, AD is the hypotenuse, and its length is also 13.

Since X1D and X2D are perpendicular to AD, we can consider ΔADX1 and ΔADX2 as two right-angled triangles sharing the same hypotenuse AD.

Triangle ΔADXE1 is similar to triangle ΔADX1, and triangle ΔADXE2 is similar to triangle ΔADX2. So, we can write the following proportion:

XE1/X1D = X2D/DE2

Since X1D = X2D (they are distances in the same rectangle), we can simplify the equation to:

XE1 = DE2

Therefore, XE1 and DE2 are equal.

Now, we can focus on triangle ΔX1DE2. This is an isosceles right-angled triangle, because X1D and X2D are equal in length, and X1E1 is equal in length to DE2.

Let's label the distance X1E1 (or DE2) as 'd'. So, to find d^2, we need to determine the value of d.

Using Pythagoras' theorem in triangle ΔX1DE2, we have:

d^2 + d^2 = (2d)^2
2d^2 = 4d^2
d^2 = (4d^2)/2
d^2 = 2d^2

Therefore, d^2 is equal to 2d^2.