Problem & Solution (3): Draw a square ABCD with given vertext A, point M on BC, point N on CD

Problem: Construct a square ABCD given point A and two points M ∈ BC, N ∈ CD

Solution:

Suppose we have constructed the square ABCD as required. Connect the points A, M and N. Observe figure 1, we can draw the following comments :

figure 1

1) ∠BCD = π / 2 = > ∠MCN = π/2, where M, N  are known. Thus vertex C is on the circle C1 of diameter MN and is on the opposite side of the vertex A through MN.

2) Through A, construct a straight line perpendicular to AN and cuts the straight line CB at P. Consider the triangles ABP and DNA: AB = AD, ∠ADN  = ∠ABP = π/2, ∠DAN = ∠BAP = > ΔADN = ΔABP => AP = AN => point P is determined. So, C is an intersection of PM with a circle C1 (C should in the opposit side compared to point A through line MN )

3) ∠MAN ≤ BAD = π/2

From the above analysis we can determine the square ABCD and point C as follows:

figure 2

Step 1: Construct the line segment MN, construct the circle C1 of diameter MN ( easily done with basic constructing skills )

Step 2: Construct the line segment NA and a straight line perpendicular to AN at A , get point P at the same plane side with point M ( through the straight lineAN ) so that AP = AN .

Step 3: Construct the straight line (d1) passing through P and M, which cuts circle C1 at C (a different point from M). Also construct the straight line (d2) through C and N. Through A, construct the straight lines perpendicular to (d1), (d2) at B and D respectively.

ABCD is the square we need to construct (see figure 2)

Proof: According to how we construct, the four vertices A,B,C, D are all square angles = > ABCD is a rectangle ( * ) . Considering the triangles ABP and DNA : AN = AP ( as above ) , ADN = ∠ABP = π/2; ∠DAN = BAP =>  ∆ADN = ∆ABP => AD = AB (**). From ( * ) and ( ** ) we have ABCD is a square .

Justification :

a) The problem will have no solution if ∠MAN > π/2, or

b) or point C lies on the same plane side with point A through line MN (including the case such that C ≡ M. This case occurs when AP ≧ AP' with  P' = AP ⋂ NM <=> ∠ANM = ∠ANP' ≦ ∠ANP = π/4 or P' = AP ⋂ NM <=> ∠ANM = ∠ANP' ≦ ∠ANP = π/4 <=> ∠ANM ≦ π/4 or ∠AMN ≦ π/4  (see figure 3)

figure 3

c) In the remaining cases the problem has a unique solution.

4.2.2014

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