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The diagram shows quadrilateral ABCD and the bisectors of the angles at A, B, C and D - HSC - SSCE Mathematics Extension 1 - Question 14 - 2018 - Paper 1
Question 14
The diagram shows quadrilateral ABCD and the bisectors of the angles at A, B, C and D. The bisectors at A and B intersect at the point P. The bisectors at A and D me... show full transcript
Worked Solution & Example Answer:The diagram shows quadrilateral ABCD and the bisectors of the angles at A, B, C and D - HSC - SSCE Mathematics Extension 1 - Question 14 - 2018 - Paper 1
Step 1
Show that PQRS is a cyclic quadrilateral
Answer
To prove that quadrilateral PQRS is cyclic, we need to show that the opposite angles are supplementary.
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Angle Relationships: Since the lines PQ, PR, QS, and SR are angle bisectors, we can express the angles at points P, Q, R, and S using the properties of angle bisectors.
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Using Angle Bisector Properties: Let angle A, B, C, and D be represented as follows:
- Set angle A as , angle B as , angle C as , and angle D as . Hence:
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Opposite Angles: It must hold that: and
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Conclusion: Since both pairs of opposite angles are supplementary, we conclude that quadrilateral PQRS is cyclic.
Step 2
By considering the expansions of $(1 + (1 + x)^n)$ and $(2 + y^r)$, show that
Answer
The expression expands to involve combinations where:
inom{n}{r} = rac{n!}{r!(n-r)!}
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Expanding the First Expression: The binomial expansion of leads to: (1 + x)^n = inom{n}{0} + inom{n}{1} (1 + x) + inom{n}{2} (1 + x)^2 + ... From this, we can derive coefficients for specific powers of x.
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Expanding the Second Expression: For , the terms also give rise to combinations.
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Setting Both Expansions Equal: By comparing both expansions, we can find the coefficients of that yield the form as requested.
This set of relations and coefficients leads to the final required combinatorial results.
Step 3
There are 23 people who have applied to be selected for a committee of 4 people.
Answer
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Selecting 4 from 23: The number of ways to choose 4 people from a pool of 23 is given by the combination formula: inom{n}{r} = rac{n!}{r!(n-r)!} Thus, for 23 applicants: inom{23}{4} = rac{23!}{4!(23-4)!} = rac{23!}{4!19!}
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Calculating the Combination: This results in: inom{23}{4} = rac{23 imes 22 imes 21 imes 20}{4 imes 3 imes 2 imes 1} = 8855
Therefore, there are 8855 different ways to form a committee of 4 people from the group of 23 applicants.