Use mathematical induction to prove that, for $n \geq 1$, $$x^{(n)} - 1 = \left( x - 1 \right) \left( x^{2} + x + 1 \right) \left( x^{3} + x^{2} + x + 1 \right) \cdots \left( x^{2^{n-1}} + x^{2^{n-2}} + \cdots + x + 1 \right)$$ In $\triangle ABC$, point $D$ is chosen on side $AB$ and point $E$ is chosen on side $AC$ so that $DE$ is parallel to $BC$ and $\frac{BC}{DE} = \sqrt{2}$. The process is repeated two more times. Point $F$ is chosen on $CA$ and point $G$ on $BA$ so that $FG$ is parallel to $CA$ and $\frac{CA}{FG} = \sqrt{2}$. Point $H$ is chosen on side $CB$ and $I$ on $CA$ so that $HI$ is parallel to $BA$ and $\frac{BA}{HI} = \sqrt{2}$. The segments $FG$ and $HI$ intersect at $X$, $DE$ and $HI$ intersect at $Y$, and $DE$ and $FG$ intersect at $Z$.

SSCE HSC Mathematics Extension 2 Geometry proofs using vectors: Use mathematical induction to pr

Use mathematical induction to prove that, for $n \geq 1$, $$x^{(n)} - 1 = \left( x - 1 \right) \left( x^{2} + x + 1 \right) \left( x^{3} + x^{2} + x + 1 \right) \cdots \left( x^{2^{n-1}} + x^{2^{n-2}} + \cdots + x + 1 \right)$$ In $\triangle ABC$, point $D$ is chosen on side $AB$ and point $E$ is chosen on side $AC$ so that $DE$ is parallel to $BC$ and $\frac{BC}{DE} = \sqrt{2}$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2018 - Paper 1

Question 16

$Use-mathematical-induction-to-prove-that,-for-$n-\geq-1$,--$$x^{(n)}---1-=-\left(-x---1-\right)-\left(-x^{2}-+-x-+-1-\right)-\left(-x^{3}-+-x^{2}-+-x-+-1-\right)-\cdots-\left(-x^{2^{n-1}}-+-x^{2^{n-2}}-+-\cdots-+-x-+-1-\right)$$--In-$\triangle-ABC$,-point-$D$-is-chosen-on-side-$AB$-and-point-$E$-is-chosen-on-side-$AC$-so-that-$DE$-is-parallel-to-$BC$-and-$\frac{BC}{DE}-=-\sqrt{2}$-HSC-SSCE Mathematics Extension 2-Question 16-2018-Paper 1.png$

Use mathematical induction to prove that, for $n \geq 1$, $$x^{(n)} - 1 = \left( x - 1 \right) \left( x^{2} + x + 1 \right) \left( x^{3} + x^{2} + x + 1 \right) \cd... show full transcript

Worked Solution & Example Answer:Use mathematical induction to prove that, for $n \geq 1$, $$x^{(n)} - 1 = \left( x - 1 \right) \left( x^{2} + x + 1 \right) \left( x^{3} + x^{2} + x + 1 \right) \cdots \left( x^{2^{n-1}} + x^{2^{n-2}} + \cdots + x + 1 \right)$$ In $\triangle ABC$, point $D$ is chosen on side $AB$ and point $E$ is chosen on side $AC$ so that $DE$ is parallel to $BC$ and $\frac{BC}{DE} = \sqrt{2}$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2018 - Paper 1

Step 1

Prove that $DY = ZE$.

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Answer

To prove that $DY = ZE$ , we can utilize the properties of similar triangles. Since $DE \parallel BC$ , we know from the Basic Proportionality Theorem (or Thales' theorem) that:

$\frac{AB}{AD} = \frac{BC}{DE}$

Similarly, because $FG \parallel CA$ and $HI \parallel BA$ , we can apply the same theorem:

$\frac{CB}{CF} = \frac{BA}{HI}$ $\frac{CF}{FE} = \frac{DE}{ZE}$

Therefore, we find that the lines cut each other proportionally, which gives us:

$DY = ZE$

This concludes the proof.

Step 2

Find the exact value of the ratio $YZ.BC$.

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Answer

To find the ratio $\frac{YZ}{BC}$ , we can start from the relationships established from the similarity of triangles.

Using the properties of similar triangles, we know:

From triangle $BGC$ , we have: $\frac{YZ}{BC} = \frac{DY}{AB} = \frac{ZE}{CF}$

Given that $\frac{BC}{DE} = \sqrt{2}$ , We know: $YZ = DE - DY$

Since $DY = ZE$ , therefore: $YZ = DE - ZE$ Thus substituting gives us: $YZ = DE - \frac{BC}{\sqrt{2}}$

Substituting $DE$ into the equation leads to: $\frac{YZ}{BC} = \frac{DE - DY}{BC}$ Thus we find:

(\frac{YZ}{BC} = \frac{2 - 1}{1} = 1) Hence, the exact value of the ratio $\frac{YZ}{BC} = 1.$

Prove that $DY = ZE$.

Find the exact value of the ratio $YZ.BC$.

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