### Roaming Rhombus

We have four rods of equal lengths hinged at their endpoints to form a rhombus ABCD. Keeping AB fixed we allow CD to take all possible positions in the plane. What is the locus (or path) of the point D?

The points P, Q, R and S are the midpoints of the edges of a convex quadrilateral. What do you notice about the quadrilateral PQRS as the convex quadrilateral changes?

The points P, Q, R and S are the midpoints of the edges of a non-convex quadrilateral.What do you notice about the quadrilateral PQRS and its area?

# Similarly So

##### Stage: 4 Challenge Level:

Correct and quite different solutions were received from Natasha Kholgade (Indian Language School, Lagos) and Julia Collins (Langley Park School for Girls, England). I have included both of them below. There is a more concise geometrical proof not involving trigonometry for anyone who wants to have another go at this problem but it still depends on the vital insight, noted by Natasha, that APQD is a cyclic quadrilateral.

Natasha's solution:

Given: ABCD is a Square, P is the mid-point of AB and DQ is perpendicular to PC.

Construction: Construct line PR parallel to AD. Join BR.

Since BP ll CR and BC ll PR (both ll AD), BPRC is a rectangle and BP = CR Thus AP = BP = CR. Also AP ll CR. Hence APCR is a parallelogram.

Hence PC ll AR. PR is a transversal. Since alternate interior angles are equal,

angleAQP = angleRAQ ?(i)
angleRPQ = angleARP ?(ii)

Given angleCQD = 90 o and anglePQD = 90 o . Thus angleCQD + angle PQD = 180 o and PADQ is a cyclic quadrilateral.

Angles in same segment are equal.

Hence angleAQP = angleARP ?(iii)
From (i) and (iii), angleRAQ = angleARP ?(iv)
From (ii) and (iii), angleRPQ = angleAQP ?(v)

Sides opposite equal angles of a triangle are equal.

Hence in triangle POQ, OP = OQ [from (v)] ?(vi)
and in triangleAOR, OR = OA [from (iv)]. ?(vii)

Adding (vi) and (vii), we get

OP + OR = OQ + OA

Or PR = AQ

Since PR ll AD (by construction) and AP ll RD, APRD is a rectangle.

Therefore AD = PR = AQ

Julia's Solution

Let angle $BPC = \alpha$ and the side of the square = $a$

In triangle $BPC$, angle $BCP = 90 - \alpha$

In triangle $QDC$, angle $DCQ = \alpha (90 - \mbox{angle} BCP)$

In triangle $PBC$, $BP = \alpha/2$, $BC = (\alpha\sqrt{5})/2$, $\cos = 1/\sqrt{5}$

In triangle $QCD$,

$QC = a \cos\alpha = a/\sqrt{5}$

$PQ = PC - QC = (a\sqrt5)/2 - a/\sqrt{5} = (3a\sqrt{5})/10$

In triangle $AQD$ - using cosine rule: $$AQ^2 = AP^2 + PQ^2 - 2xAPxPQ \cos(180 - \alpha)$$ $$\quad= a^2/4 + 9a^2/20 - 3a^2\sqrt{5}/10(-\cos\alpha)$$ $$\quad = a^2/4 + 9a^2/20 - 3a^2\sqrt{5}/10(1/\sqrt{5})$$ $$\quad= a^2$$

$AQ = AD = a$

We received the following solution that uses some circle theorems to prove the result:

Angle BPC = angle APD (P is the midpoint of AB, so triangles PAD and PBC are congruent (SAS))

Also, PADQ is a cyclic quadrilateral, because angle PAD and angle CQD are both $90^{\circ}$, and a quadrilateral whose opposite angles add up to $180^{\circ}$ is cyclic.

Thus, angle ADQ = angle BPC, (because angle APQ + angle ADQ add up to $180^{\circ}$, and angles APQ and BPC are on a straight line so also add up to $180^{\circ}$

But angle BPC = angle APD, so angle ADQ = angle APD.
Angle APD and angle AQD are in the same segment so they are equal.

Therefore, triangle AQD is an isosceles triangle, so AQ = AD.