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Cosines Rule

Three points A, B and C lie in this order on a line, and P is any point in the plane. Use the Cosine Rule to prove the following statement.

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DOTS Division

Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.

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Root to Poly

Find the polynomial p(x) with integer coefficients such that one solution of the equation p(x)=0 is $1+\sqrt 2+\sqrt 3$.

Matchless

Stage: 3 and 4 Challenge Level: Challenge Level:2 Challenge Level:2

Well done to all the people who combined the expressions in such a way as to eliminate the $x$ or $y$ variable. For example Sakshi from Singapore did this :

$(2x + 3y - 20) + (4x + 5y - 72) = 6x + 8y - 92 $
and
$(5x - 2y + 38) + (x - 4y + 108) = 6x - 6y + 146 $
These two results must both equal the same value.
And since $6x$ is in both expressions, contributing the same to each expression whatever its actual value, $8y - 92$ must match $146 - 6y $
from which it follows that $14y = 238$ with $y = 17 $

x can then be found with a similar approach :
(2x + 3y - 20) + (5x - 2y + 38) = 7x + y + 18
(4x + 5y - 72) + (x - 4y + 108) = 5x + y + 36
And since 7x + y + 18 matches the value of 5x + y + 36 , x must be 9

Jennifer from Our Lady's Convent had another good approach :

Set any two of the five expressions equal to each other.
$2x + 3y - 20 = 5x - 2y + 38 $
and then simplify the expression to get :
$-3x + 5y - 58 = 0 $

Set any other pair of expressions equal to each other
$4x + 5y - 72 = 3x - y + 39 $
then again simplifyingto get :
$x + 6y - 111 = 0 $

I now have two simultaneous equations, and as there are two unknown values in each, I will make $x$ the subject of one of them, and substitute it into the other.

Substitute $x = 111 - 6y$ into $-3x + 5y - 58 = 0$
$-3(111 - 6y) + 5y - 58 = 0 $
$-333 + 18y + 5y - 58 = 0 $

$23y = 391$ and so $y = 17$

Substitute $y = 17$ into $-3x + 5y - 58 = 0$
to get : $-3x + (5 x 17) - 58 = 0 $
so $x = 9$

Jennifer then checked the whole thing by substituting $y = 17$ and $x = 9$ into all expressions to find that for that combination of values each expression produced $49$ . Brilliant!

Alice from Colyton Grammar School explains that in fact only $3$ of the expressions are needed, any three will do. Here's her method :

To find the values of $x$ and $y$ for all the expressions to be equal let: $2x + 3y - 20 = 5x - 2y + 38$
simplify to get $3x - 5y = -58$ ......($1$)
similarly let: $4x + 5y - 72 = 5x - 2y + 38 $
simplify to get $x - 7y = -110$ ......($2$)
solve ($1$) and ($2$) simultaneously
$(1) - 3*(2)$ leads to $16y = 272$ so that $y = 17$
Sub $y = 17$ into ($2$)
$x = -110 + 7y = 9 $

So only $3$ equations are needed to solve the problem as only two equations are needed to solve an equation with two variables however, the additional equations given can be used to check if the $x$ and $y$ values are correct.

A teacher's account of working with students on this problem has been added to the Notes section, do have a look.