Category Archives: STEP and MAT

STEP 2/2009/8

The non-collinear points A, B, C have position vectors a, b, c respectively. The points P and Q have position vectors p and q, respectively, given by

p = λ a + (1 − λ)b

q = μ a + (1 − μ)c

where 0 < λ < 1. Draw a diagram showing A, B, C, P and Q

Given that CQ.BP = AB.AC, find μ in terms of λ, and show that, for all values of λ, the line PQ passes through the fixed point D, with position vector d given by

d = − a + b + c

What can be said about the quadrilateral ABDC?

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STEP 2/2008/8

The points A and B have position vectors a and b, respectively, relative to the origin O. The points A, B, and O are not collinear. The point P lies on AB between A and B such that

AP : PB = (1-\lambda):\lambda

Write down the position vector of P in terms of a, b, and λ

Given that OP bisects the angle AOB, determine λ in terms of a and b, where

a = |a| and b = |b|

The point Q also lies on AB between A and B, and is such that AP=BQ. Prove that

OQ2 − OP2 = (b-a)2

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STEP II/1996/3

The Fibonacci numbers Fn are defined by the conditions: F0=0, F1=1, and Fn+1 = Fn + Fn-1 for all n ≥ 1. Show that F2=1, F3=2, F4=3, and compute F5, F6 and F7.

Compute Fn+1Fn-1 − F02 for a few values of n; guess a general formula and prove it by induction, or otherwise.

By induction on k, or otherwise, show that Fn+k=FkFn+1+Fk-1Fn for all positive integers n and k.

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