UKMT Senior Maths Challenge 2019

Thought this was a tough paper

[CrazeZombsRoyaleioGamer]

the way u explained the questions is amazing, I got the senior math challenge 2022 next tuesday hopefully I do well

[dontspam]

For Q16, you cld also multiply the second equation by 2 and then subtract it from the 1st to get x+y+z=30 which implies mean is 10

[mahinkhadir]

For question 25 35:13, a much easier way would have been if you have extended the line of length 0.5 to the side of the square. Labelling the sides (if we take side length of square as s) would make it obvious that pythagoras can be applied, and it would result in a quadratic in terms where you solve for s. The equation should look like this 1^2 = (0.5s)^2 + (0.5+s)^2. It is a right angled triangle because the figure is symmetrical about the line 0.5 + s

[blatser]

For question 12, because the height is 2, can we not create a right angle triangle on the left side of the rhombus, with angle 60 and opposite = 2 to work out the side length of the rhombus and because all lengths on a rhombus are equal, the base is equal to the hypotenuse of the triangle we created?

[darainsyed]

For q20 (24:40) I think there's an easier way to find the angle instead of finding all angles on the quad

Let angle(ZXY) = x
As you have shown angle(XOP) = 180 - 2x
As you have shown angle(XYZ) = (180 - x)/2 so simplifies to 90 - x/2

AOY makes a triangle itself therefore angle(AOY) = 180-(60 + 90 - x/2) = 30 + x/2
As we know angles on a straight line = 180

angle(AOY) + angle(XOP) = 180

30 + x/2 + 180 - 2x = 180
3/2x = 30
x = 20

[Dz-iunk]

For Q23, I actually flattened out the squares into a two dimensional shape and then calculated the area which turned out to be a trapezium🤣

[tosynariyo]

At 18:35, surely a more elegant way to get the answer is multiply the second equation by 2 and take the first one away from that to get x+y+z=30. So mean is 10 😊