JEE Mathematics Geometrical Constructions Notes Set 03

Mean Proportional by Construction

The mean proportional (geometric mean) of two lengths a and b is the length x with \( x^2=ab \). To construct it, lay a and b end to end on a line so that BD = a and DC = b, draw a semicircle on BC as diameter, and erect the perpendicular at D up to the semicircle at A. Then DA is the required length.

Why it works: angle BAC is a right angle (angle in a semicircle), and AD is the altitude to the hypotenuse, so \( AD^2=BD\cdot DC \). The legs satisfy \( AB^2=BD\cdot BC \) and \( AC^2=DC\cdot BC \). Example: with BD = 7 and DC = 5 we get \( AD=\sqrt{35}\approx5.92 \) cm.

Facts About Circles Used in Constructions

1. The angle in a semicircle is a right angle.
2. The angle at the centre is twice the angle at the circumference on the same arc.
3. Angles in the same segment are equal.
4. The angle between a tangent and a chord equals the angle in the alternate segment.
5. A tangent is perpendicular to the radius at the point of contact, and two tangents from an external point are equal.
6. For a chord of length a that subtends angle A at the circumference, \( a=2R\sin A \); the distance of the centre from the chord is \( R\cos A \) (for acute A).

Triangle with a Given Base, Vertex Angle and One More Condition

The vertex opposite the base lies on a circular arc that sees the base at the given angle. The extra condition then cuts this arc at the vertex.

How to draw the arc: at one end B of the base draw the line BX making angle A with the base. Draw BY perpendicular to BX and the perpendicular bisector of the base; they meet at the centre O. The circle with centre O and radius OB is the required circle, and by the tangent-chord theorem the angle in the alternate segment equals the angle A.

Worked check: BC = 7 cm, angle A = 70 degrees. \( R=\frac{7}{2\sin70^\circ}\approx3.73 \) cm, and the centre is \( R\cos70^\circ\approx1.27 \) cm from BC. With the foot of the altitude 4.5 cm from B, \( AD\approx4.86 \) cm. For AB = 5.4 cm and angle C = 65 degrees, R is about 2.98 cm, so a median between about 1.72 cm and 4.24 cm gives a triangle; the median 3.7 cm gives two mirror-image triangles.

Constructing Tangents

Tangents from an external point P to a circle with centre O: join OP, draw the circle on OP as diameter, and join P to the points A and B where the two circles meet. Because angle OAP is an angle in a semicircle, PA is perpendicular to OA, so PA and PB touch the circle. Length: \( PA=PB=\sqrt{OP^2-r^2} \). For OP = 7 cm and r = 3 cm this is \( \sqrt{40}\approx6.32 \) cm. The angle between the tangents is \( 2\sin^{-1}(r/OP) \).

Common Tangents to Two Circles

For circles with centres P and Q at distance d and radii \( r_1 > r_2 \):

Draw tangents from the other centre Q to the auxiliary circle, then shift them outward by \( r_2 \) (direct) or across by \( r_2 \) (transverse) using parallels. Example: r = 3 and 2, d = 7: direct length \( \sqrt{48}\approx6.93 \) cm and indirect length \( \sqrt{24}\approx4.90 \) cm.

Regular Polygons in a Circle

A regular n-gon inscribed in a circle of radius R has side \( 2R\sin(180^\circ/n) \). Equilateral triangle: \( \sqrt3R \). Square: \( \sqrt2R \). Regular hexagon: R, so six chords equal to the radius close the circle.