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The School of Athens
Euclid, as imagined by Raphael in this detail from The School of Athens.
Jamshed Al-Kashi
Al-Kashi was one of the greatest mathematicians of the 15th century, calculating the sin(1*) correct to 16 decimal places.
Al-Beruni
The 11th century scientist and philosopher managed to estimate the radius of the earth using a very creative method that employed trigonometry.
54 comments:
Any ideas for this problem? I've gotten that triangles ABY, CAX, and BCZ are all equal, and all 1/3 of triangle ABC but don't know where to go.
Sam, I don't know if this helps at all, but if you look at the three triangles you mentioned or traingles AZB, CYA, and XBC aren't they all made up of the "same" parts? I just remember a problem we did in class that had to do with subtracting equal parts and such, so I don't know if that applies to this or not. just thinking...
I don't know why I hadn't seen it sooner, but Sam, if you look at triangle BPY inside of triangle BCZ, you'll see that if you flip BPY, it's 1/3 of BCZ. You can prove their similarity if you've already gone through and proved each group of triangles congruent.
I lied. Not 1/3; 1/9. Sorry. I just went through and realized my typo.
Well, perhaps I was wrong again or I took a wrong turn somewhere. After subtracting all the parts from triangle ABC, I got triangle PQR to be 1/9 of ABC, when we're supposed to prove it to be 1/7. I used to do this on some homework problems as well, but my notes are unclear as to how to solve it. Any tips?
I have a suggestion. Whitney, I looked at BPY and it gave me a push in the right direction. I figured out what type of triangle PQR was, and from there I could play with some congruence of triangles. I haven't gone through and done the actual numbers yet, so this is only a suggestion.
ALSO! I was wondering if this was a typo:
In question #4, it says "X is on AB and AX : AB = 1 : 2"
Should it not be "X is on AB and AX : XB = 1 : 2"
Just wondering...
Yes Emily, you caught the typo. I had announced it in class on Friday. Indeed, AX:XB = 1:2
I keep getting caught up after the point that Emily raised in regards to the type of triangle PQR is. I have found out which type of triangle that is, but I'm now stuck. Can anyone give me a push in the right direction?
I found a relationship between triangle BPY and triangle QPR, but I'm having trouble with the quadrilaterals such as PRCY. I can't find any ratios that are helpful. I was thinking you might be able to divide ABC into triangle QCY and two other triangles or something... I'm just trying to throw some ideas out there...
whitney i did the exact same thing as you but i'm not sure how we messed it up.
I've reached the same conclusion as whitney and amanda. I'm wondering where we're taking a wrong turn...
i'm not sure but did you guys maybe prove that oqr is similar to abc by a ratio of 3:1? then if you square the area that would give you that pqr is 1/9 of abc.
i am stuck like steven after i found out what kind of triangle pqr is and proved alot of congruence stuff. does anyone have any ideas of where to go next?
I am still confused about how ABY, CAX, and BCZ are all equal? I think I am totally missing something.
Slaughter- It is given that triangle ABC is equilateral, so all sides and altitudes of ABC are equal.
The first thing I did was prove each set of triangles congruent, such as ABY, BCZ, and CZX; BYP, CZR, and AXQ, and then I proved that PQR is an equilateral triangle because all of its angles were equal. Then I went ahead and looked at the triangles ABY, BCZ, and CZX themselves and found that they were each 1/3 of ABC, and then the smaller triangles BYP, CZR, and AXQ are 1/9 of that, but maybe I messed something up. Going through this, I keep getting that PQR is 1/9 the area of ABC. Help?
How do you prove that ABY, CAX, and BCZ are all 1/3 of triangle ABC?
I,too, have found what RQP is, and was wondering if someone could give me a hint as to where to go from there. I have tried looking at other triangles, and other ratios, but I seem to be in a math block right now.
Slaughter- find the heights and bases of those triangles
I added two lines parallel to BZ and if I can prove that they cut transversals into equal segments, then I can show that triangle PQR is 3/21st=1/7th of triangle ABC.
Scott. Thanks for your post. I see what you are saying, but where did you get the 21 from?
I got 21 because i divided everything into triangle AQX size pieces, but I still haven't found a way to prove that my auxillary lines are parallel and cut the transversals in a 1:1 ratio.
Scott, aren't your auxiliary lines constructed parallel?
Emily- Well, if I do that, then I can't prove that they cut transversals in a 1:1 ratio and vice versa. I need to be able to prove both conditions for my proof to work.
Scott- maybe I am not seeing what you are explaining, but why can't you construct the parallel lines and then say they cut the transversals in a 1:1:1 ratio?
Is there any way anyone can elaborate on Scott's idea. I am having trouble visualizing even the drawing of the parallel lines to BZ. What's the significance of the three proportional segments created on the transversals (which are??), in theory, by these parallel lines. If anyone could take a minute to help me, it would be most appreciated. Thanks
I could do that, but I need one of the parallel lines to pass through Q. If I say that they are parallel and cut transversals in a 1:1 ratio, then I can't say that it goes through Q also. My idea is to prove that the altitude from Q for triangle QRP is equal to the altitude of CRPY from point C.
I drew this to help me visualize what Scott is saying (Vranian, this might help you?)
http://i41.tinypic.com/2eqgpyp.png
Click the link.
Emily- the one through Q is right, and my other one was through C. I don't think you need the other two. Ideally, the parallel line through Q also intercepts AC at a point so that the parallel lines trisect AC.
I see, so we're looking at something similar to this:
http://i44.tinypic.com/2rxcy14.png
I can't click the link
Try copying and pasting it into your address bar in your internet application.
Yeah, basically that.
Wait, Scott, so for you to be saying that AQX is 1/21 ABC, then you proved that it was 1/7 of ABY, which is what I had been trying to do. How did you do that? If I could do that, then I think I'd be able to prove that PQR is 1/7 ABC.
i'm confused how does drawing those parallel lines help prove that pqr is 1/7 abc? oh and thanks for the diagram emily
Does anyone have a complete proof for this problem?
If so, could you give some hints because I am still stuck on trying to prove that ABY, CAX, and BCZ are all 1/3 of triangle ABC
scott- i still am a little confused, but can you say that your line through q bisects AZ? then you have a 1:1:1 ratio, so wouldn't the three lines therefore have to be parallel because they divide AC into equal segments? does saying that assume too much?
Whitney- That was only a result of proving that PRQ is 1/7 of ABC. If you could prove that first, then you could prove it, but I found that through my parallel line approach (which I haven't proved yet).
Slaughter- Area of triangle=1/2 base x height. So you need to find their bases and altitudes, which should be equal because triangle ABC is equilateral.
Virginia- I think that is assuming too much because the (parallel) lines need to intersect 2 lines into proportional segments to be parallel.
Has anyone proven that CR=RQ? if we could prove that, then the three lines would intersect 2 tranversals proportionally and therefore be parallel. (and everything would work out)
and has anyone proved it using another approach?
i'm wondering if there's any way to use similar triangles to prove it because we were given the ratios of the sides- any ideas?
I'm really close to proving it. I don't think that the parallel lines are necessary, but they did get me thinking about this. It's hard to explain, but its a quadrilateral QZC(the fourth vertex is 1/3 of the distance of RP on R). I'm trying to prove that CQ bisects the other diagonal.
I bet this whole problem is really a simple proof, and there is some way to prove it in like ten minutes.
I got it. You don't need the parallel lines. I proved that two of them (in the quadrilateral I mentioned in my last post) were congruent, thus proving that CR=RQ.
So you proved that CQ bisects BZ in the quadrilateral you mentioned? How did this help you?
I see exactly what you are saying, Scott. But as Victoria is questioning, how does that help you exactly?
Once you prove that CZR and the other triangle in the quadrilateral with Q and R as two of the vertices, are congruent, then you can divide CRPY up into triangles of equal bases because you know the ratio of ZR to ZP.
I have a question. What if... you could divide CB into 7 equal segments? If you were to figure that out in proportion to the area of triangle PBY (Which I've done) , you should find that 3 PBY's is 3/21 or 1/7. However, the question is, can you split CB into 7 equal segments? I'm doubtful this will work, but it's worth an investigation.
SCRATCH THAT. 3 PBY's do not make up triangle PQR. I lied. Haha
Scott, I don't mean to be a pest, but I still don't see how your proof works. Could you explain just a little of it and then maybe I can begin to see it?
What is the quadralateral that you are talking about?
The quadrilateral is QZC and then the 4th point is 1/3 of the the line RP closer to point R...
I'm not sure where you are lost, but the concept is to divide up the triangle into little triangles of area equal to ZRC. Its hard to explain online, but using equal angles that I found, I proved that the quadrilateral that Steven just posted is a parallelogram, and then you know lots of new ratios and lengths of the sides and diagonals which can be used to break up triangle ABC into smaller triangles.
Scott, I know exactly what you mean now, but I'm having issues proving that QZCO (that's what I named the point) is a parallelogram, or at least that QR=CR. Halp? I know it's really late... But I also know that if we prove that QR=CR, then we know through the transitive property that PR=BP, and then BZ will be split into sevenths.
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