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Maths question help please

Hi guys, I get up to k=6 and then I'm confused about "If P, Q, and R lie on the circle, and triangle PQR is equilateral, write down the coordinates of the two vertices Q and R" bit. Also, I don't understand the mark scheme answer for that bit as well. Can someone please help? Many thanks
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Original post by sienna2266
Hi guys, I get up to k=6 and then I'm confused about "If P, Q, and R lie on the circle, and triangle PQR is equilateral, write down the coordinates of the two vertices Q and R" bit. Also, I don't understand the mark scheme answer for that bit as well. Can someone please help? Many thanks
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The solution in the mark scheme is a bit of a 'trick-spot' - they just notice that the angle given is 60 degrees, and then it is geometrically clear that the second vertex is given by reflecting the point in the x-axis and then the location of the third vertex follows easily, again by geometric intuition.

To see this, draw an equilateral triangle, and mark its central point (which in this question is also the centre of the circle). Draw lines from the vertices to the central point, and calculate what the angles in the 3 smaller triangles are that result.
(edited 6 years ago)
Original post by sienna2266
Hi guys, I get up to k=6 and then I'm confused about "If P, Q, and R lie on the circle, and triangle PQR is equilateral, write down the coordinates of the two vertices Q and R" bit. Also, I don't understand the mark scheme answer for that bit as well. Can someone please help? Many thanks


In an equilateral triangle, all vertices are spaces 120 degrees apart about its origin. You found one point, and note that its angle to the horizontal is 60 degrees. This means if you go another 60 degrees you'll end up at the point where another vertex must be. But if you go those another 60 degrees, you'll end up at a point that's just a reflection of the point P in the x-axis.
Then the other point can easily be deduced since we can establish that it must lie on -ve x-axis, which is precisely the point (-6,0) on the circle.
(edited 6 years ago)
Reply 3
Original post by RDKGames
In an equilateral triangle, all vertices are spaces 120 degrees apart about its origin. You found one point, and note that its angle to the horizontal is 60 degrees. This means if you go another 60 degrees you'll end up at the point where another vertex must be. But if you go those another 60 degrees, you'll end up at a point that's just a reflection of the point P in the x-axis.
Then the other point can easily be deduced since we can establish that it must lie on -ve x-axis, which is precisely the point (-6,0) on the circle.


Thanks very much! :smile:
Reply 4
Original post by RDKGames
In an equilateral triangle, all vertices are spaces 120 degrees apart about its origin. You found one point, and note that its angle to the horizontal is 60 degrees. This means if you go another 60 degrees you'll end up at the point where another vertex must be. But if you go those another 60 degrees, you'll end up at a point that's just a reflection of the point P in the x-axis.
Then the other point can easily be deduced since we can establish that it must lie on -ve x-axis, which is precisely the point (-6,0) on the circle.


"...You found one point, and note that its angle to the horizontal is 60 degrees" - how do you know the angle is 60 degrees here? Many thanks
Original post by sienna2266
"...You found one point, and note that its angle to the horizontal is 60 degrees" - how do you know the angle is 60 degrees here? Many thanks


It's a right-angled triangle whose all 3 sides you know. Just use trig to figure it out.
Reply 6
Original post by RDKGames
It's a right-angled triangle whose all 3 sides you know. Just use trig to figure it out.


Ahh ok I understand this. Just to check with something else...
For a point of an equilateral triangle,is its angle to the horizontal always going to be 60 degrees? I think it might be different for different coordinates? For example, if P was (4,1) instead of (3,root 27), the angle P (4,1) makes to the horizontal is 14 degrees. This means if you go another 14 degrees you'll end up at the point where another vertex must be? But if you go "those another 14 degrees" you'll end up at a point that's just a reflection of the point P in the x axis? And then how do you establish the other point? Many thanks
(edited 6 years ago)
Original post by sienna2266
Ahh ok I understand this. Just to check with something else...
For a point of an equilateral triangle,is its angle to the horizontal always going to be 60 degrees?


No. The reflection only works because the angle of inclination is half of the required amount. If it's a different configuration then more work needs to be done to establish other points.
Reply 8
Original post by RDKGames
No. The reflection only works because the angle of inclination is half of the required amount. If it's a different configuration then more work needs to be done to establish other points.


So for a known point of an equilateral triangle, if its angle to the horizontal is not 60 degrees then you cannot carry out the reflection thing to get the other point because all vertices should be spaced 120 degrees apart about its origin in an equilateral triangle?

And by different configuration, do you mean different angle?

Many thanks




(edited 6 years ago)
Original post by sienna2266
So for a known point of an equilateral triangle, if its angle to the horizontal is not 60 degrees then you cannot carry out the reflection thing to get the other point because all vertices are spaces 120 degrees apart about its origin in an equilateral triangle?

And by different configuration, do you mean different angle?


The angle doesn't have to be to the horizontal. It can be 60 degrees to the vertical and you can still carry out reflection in the y-axis for a second vertex.

By different configuration, I mean the vertices are anywhere on the circle as long as they're all equal distance from each other. There are plenty more situations where the approach to find them isn't as nice as it is in this question due to their configurations.
Original post by RDKGames
The angle doesn't have to be to the horizontal. It can be 60 degrees to the vertical and you can still carry out reflection in the y-axis for a second vertex.

By different configuration, I mean the vertices are anywhere on the circle as long as they're all equal distance from each other. There are plenty more situations where the approach to find them isn't as nice as it is in this question due to their configurations.


Ahh okay but specifically, for a known point of an equilateral triangle , its angle to the horizontal or vertical should be 60 degrees to be able to carry out the reflection thing in the x or y axis to get the other vertex?

Many thanks
Original post by sienna2266
Ahh okay but specifically, for a known point of an equilateral triangle , its angle to the horizontal or vertical should be 60 degrees to be able to carry out the reflection thing in the x or y axis to get the other vertex?

Many thanks


Yes, should be obvious why this must be the case.
Original post by RDKGames
Yes, should be obvious why this must be the case.


Thank you so much for helping me understand this!!😀
Original post by sienna2266
Thank you so much for helping me understand this!!😀


Low-key feel like taking your teacher's salary by the amount of questions you ask on here :ninja: all good though :tongue:
Original post by RDKGames
Low-key feel like taking your teacher's salary by the amount of questions you ask on here :ninja: all good though :tongue:


Haha. I appreciate it a lot. My teacher just goes along with whatever I think about a maths question 😂

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