Three natural numbers a, b and c, such that $a^2 + b^2 = c^2$, are called a Pythagorean triple. (i) Let $a = 2n + 1, b = 2n^2 + 2n$ and $c = 2n^2 + 2n + 1$. Pick one natural number n and verify that the corresponding values of a, b and c form a Pythagorean triple. (ii) Prove that $a = 2n + 1, b = 2n^2 + 2n$ and $c = 2n^2 + 2n + 1$, where $n \in \mathbb{N}$, will always form a Pythagorean triple. (b) $ADEC$ is a rectangle with $|AQ| = 7 m$ and $|AD| = 2 m$, as shown. $B$ is a point on $[AC]$ such that $|AB| = 5 m$. $P$ is a point on $[DE]$ such that $|DP| = x m$. (i) Let $f(x) = |PA|^2 + |PB|^2 + |PC|^2$. Show that $f(x) = 3x^2 - 24x + 86$, for $0 \leq x \leq 7, x \in \mathbb{R}$. (ii) The function $f(x)$ has a minimum value at $x = k$. Find the value of k and the minimum value of $f(x)$.

Leaving Cert Mathematics Geometry: Three natural numbers a, b and c

Three natural numbers a, b and c, such that $a^2 + b^2 = c^2$, are called a Pythagorean triple - Leaving Cert Mathematics - Question 7 - 2014

Question 7

Three natural numbers a, b and c, such that $a^2 + b^2 = c^2$, are called a Pythagorean triple. (i) Let $a = 2n + 1, b = 2n^2 + 2n$ and $c = 2n^2 + 2n + 1$. Pick o... show full transcript

Worked Solution & Example Answer:Three natural numbers a, b and c, such that $a^2 + b^2 = c^2$, are called a Pythagorean triple - Leaving Cert Mathematics - Question 7 - 2014

Step 1

Let $n = 1$: Pick one natural number n and verify that the corresponding values of a, b and c form a Pythagorean triple.

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Answer

Substituting $n = 1$ :

$a = 2(1) + 1 = 3$
$b = 2(1)^2 + 2(1) = 4$
$c = 2(1)^2 + 2(1) + 1 = 5$

Now we check:

$3^2 + 4^2 = 9 + 16 = 25 = 5^2$

Thus, $3$ , $4$ , and $5$ form a Pythagorean triple.

Step 2

Prove that $a = 2n + 1, b = 2n^2 + 2n$ and $c = 2n^2 + 2n + 1$, where $n \in \mathbb{N}$, will always form a Pythagorean triple.

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Answer

We analyze:

$a^2 = (2n + 1)^2 = 4n^2 + 4n + 1$ $b^2 = (2n^2 + 2n)^2 = 4n^4 + 8n^3 + 4n^2$ $c^2 = (2n^2 + 2n + 1)^2 = 4n^4 + 8n^3 + 5n^2 + 4n + 1$

Now we compute:

$a^2 + b^2 = (4n^2 + 4n + 1) + (4n^4 + 8n^3 + 4n^2)$ (combining terms)

This simplifies to: $4n^4 + 8n^3 + 8n^2 + 4n + 1$

Thus, we need to check: $c^2 = 4n^4 + 8n^3 + 5n^2 + 4n + 1$

Here, it shows: $a^2 + b^2 = c^2$

Therefore, $a$ , $b$ , and $c$ will always form a Pythagorean triple.

Step 3

Show that $f(x) = |PA|^2 + |PB|^2 + |PC|^2$.

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Answer

To compute the distances:

$|PM| = |DP| + |ME| = (7 - x) - 2$

Then we have: $f(x) = |PA|^2 + |PB|^2 + |PC|^2 = |PA| = |DP| + |DA| + |DA|^2 + |PB|^2 + |PC|^2$

The full expression is: $f(x) = x^2 + 2^2 + (5 - x)^2 + 2^2 + 2^2$

By simplifying we find: $f(x) = 3x^2 - 24x + 86$

Step 4

Find the value of k and the minimum value of f(x).

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Answer

To find the minimum value, we take the derivative:

$f'(x) = 6x - 24$

Setting the derivative to zero for minimization: $6x - 24 = 0 \Rightarrow x = 4 = k$

Now we evaluate: $f(4) = 3(4^2) - 24(4) + 86$ $= 48 - 96 + 86 = 38$

Thus, the minimum value of $f(x)$ is $38$ .

Three natural numbers a, b and c, such that $a^2 + b^2 = c^2$, are called a Pythagorean triple - Leaving Cert Mathematics - Question 7 - 2014

Worked Solution & Example Answer:Three natural numbers a, b and c, such that $a^2 + b^2 = c^2$, are called a Pythagorean triple - Leaving Cert Mathematics - Question 7 - 2014

Let $n = 1$: Pick one natural number n and verify that the corresponding values of a, b and c form a Pythagorean triple.

Prove that $a = 2n + 1, b = 2n^2 + 2n$ and $c = 2n^2 + 2n + 1$, where $n \in \mathbb{N}$, will always form a Pythagorean triple.

Show that $f(x) = |PA|^2 + |PB|^2 + |PC|^2$.

Find the value of k and the minimum value of f(x).

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