# Q2 bmo1 2006/07

Hi all, I have a question about Q2; how do you prove MQP and MNP have the same areas as triangles APC and AMC respectively? Do they have the same areas? Proving AMPC = 1/3 ABCD was very easy but if anything I've proven MQP is not equal to APC (which of course isn't possible). Although the base for both triangles is x, the height is different, as the line perpendicular to DC which goes through A is obviously not the same length as the line perpendicular to DC which goes through M, at least in every diagram I've drawn. The same could be said for triangles MNP and AMC; they both have y as a base, but different heights. Is there something I'm doing wrong?
Original post by another.student!
Hi all, I have a question about Q2; how do you prove MQP and MNP have the same areas as triangles APC and AMC respectively? Do they have the same areas? Proving AMPC = 1/3 ABCD was very easy but if anything I've proven MQP is not equal to APC (which of course isn't possible). Although the base for both triangles is x, the height is different, as the line perpendicular to DC which goes through A is obviously not the same length as the line perpendicular to DC which goes through M, at least in every diagram I've drawn. The same could be said for triangles MNP and AMC; they both have y as a base, but different heights. Is there something I'm doing wrong?

(edited 11 months ago)
(not sure if the photo is working? I've literally been trying to upload this for half an hour, apologies. I cannot seem to find the attach image button everyone is talking about.)

I've gotten a bit further since my last post, and realised that MNS + RQP = SPC + AMR, but I have no idea how to go about proving this (this assumes R is the point where AP cuts the line MQ, and S is the point where MC cuts the line NP).

edit: i've added the photo to a photo-sharing website so maybe you'll be able to see it now. sorry for the delay:
https://ibb.co/TKmwNw5
(edited 11 months ago)
Original post by another.student!

(not sure if the photo is working? I've literally been trying to upload this for half an hour, apologies.)

I've gotten a bit further since my last post, and realised that MNS + RQP = SPC + AMR, but I have no idea how to go about proving this.

If you click on the camera icon in the reply toolbar it should work. But if not, just upload to another site and link it maybe? From your OP and guessing your diagram you said that
MQP is not equal to APC
which Id agree with, though the area of MPQ is equal to another triangle which is useful. The question is about relating the areas of triangles to each other which is the old greek way of showing that two areas (quadrilateral) were equal.
(edited 11 months ago)
Original post by another.student!
(not sure if the photo is working? I've literally been trying to upload this for half an hour, apologies. I cannot seem to find the attach image button everyone is talking about.)

I've gotten a bit further since my last post, and realised that MNS + RQP = SPC + AMR, but I have no idea how to go about proving this (this assumes R is the point where AP cuts the line MQ, and S is the point where MC cuts the line NP).

edit: i've added the photo to a photo-sharing website so maybe you'll be able to see it now. sorry for the delay:
https://ibb.co/TKmwNw5

Thats just the diagram with little working, and not sure where youre going with the opposite angles that are marked. In proving AMCP = 1/3 ABCD, Im presuing you split AMCP into two triangles, then argued that each triangle was 1/3 of ABC and 1/3 of CDA ...?

For the second part, you could argue that the areas of AMCP and MNPQ are equal or that the latter is 1/3 of ABCD. Do you have a feeling about which would be easier?
(edited 11 months ago)
Hi, like I've said I solved the first half of the question so I just redrew the diagram to make it look a bit more neat - but I did indeed just split it with the line AC and prove it with two individual triangles.

As for the second half of the question, which is what I'm stuck on, I think it's easier to prove AMCP = MNPQ; I just need to prove that AMR + CPS = MNS + RQP, as shown in the image I've sent over, as that would therefore be the same size as AMPC (with MRPS being in both quadrilaterals), I just have no idea as to how to go about this. I just marked the opposite angles because I couldn't think of anything else.
Original post by another.student!
Hi, like I've said I solved the first half of the question so I just redrew the diagram to make it look a bit more neat - but I did indeed just split it with the line AC and prove it with two individual triangles.

As for the second half of the question, which is what I'm stuck on, I think it's easier to prove AMCP = MNPQ; I just need to prove that AMR + CPS = MNS + RQP, as shown in the image I've sent over, as that would therefore be the same size as AMPC (with MRPS being in both quadrilaterals), I just have no idea as to how to go about this. I just marked the opposite angles because I couldn't think of anything else.

Id probably think more about how you can adapt the first approach to the second case and show AMCP=MNPQ. Which diagonal did you use when you split AMCP up ...

It does help to upload all your working at the start to give a relevant hint. Also, when youre thinking about how to approach a geometry problem, try and list all the relevant appraoches/relationships that might be useful and work through them/cross off as necessary.
(edited 11 months ago)
Original post by mqb2766
Id probably think more about how you can adapt the first approach to the second case and show AMCP=MNPQ. Which diagonal did you use when you split AMCP up ...

It does help to upload all your working at the start to give a relevant hint. Also, when youre thinking about how to approach a geometry problem, try and list all the relevant appraoches/relationships that might be useful and work through them/cross off as necessary.

I used AC, would that be useful again?

Alright thanks for your tips, I'll try and re-write out all my working. I tried this method before yesterday, but (as you saw on the other thread) it just left me more confused because I proved what I thought WASN'T correct. We'll see if I get luckier this time!
Original post by another.student!
I used AC, would that be useful again?

Alright thanks for your tips, I'll try and re-write out all my working. I tried this method before yesterday, but (as you saw on the other thread) it just left me more confused because I proved what I thought WASN'T correct. We'll see if I get luckier this time!

MP is probably more useful as it occurs in both quadrilaterals?
Original post by mqb2766
MP is probably more useful as it occurs in both quadrilaterals?

I tried that but it completely went over my head to try and use it for the quad AMCP too! Thanks so much
Original post by another.student!
I tried that but it completely went over my head to try and use it for the quad AMCP too! Thanks so much

AC is the way to do the first equality, but MP is more useful for the second.

As mentioned in the previous post, its worth being explicit / sketching / ... about all decompositions/theorems/... that arise from the info in the question then thinking about which might give towards a solution. Dont just assume you'll magically spot all the constructions / arguments at the start and sometimes its a case of working backwards so saying I could prove X if I knew Y, then thinking about how to get Y ...

In this case triangles of equal area occur when the base is the same size (so any adacent intervals on one side) and the other vertex is common, so any point on the opposite side, not just at A,B or C,D.
(edited 11 months ago)
Original post by another.student!
edit: i've added the photo to a photo-sharing website so maybe you'll be able to see it now. sorry for the delay:
https://ibb.co/TKmwNw5

FWIW, TSR often doesn't deal with image attachments very well anyway - it has a tendency to ignore rotation flags (so images taken on phones display rotated) and also sometimes decides a 128 pixel wide version of your image should be enough for anybody.

So external site often works out better even if you can get an attachment to work.

(Sorry I can't help on the geometry...!)