....you can solve this math problem:
problem1.jpg
Find the area of the large triangle, in exact terms.
NOTE: The figure is not drawn to scale.
Give details on how you arrived at the answer. If you just post a number, I will not reply to your post, even if it's correct.
This question is solvable. It took me more than half a hour. It's a hard one though. However, it requires nothing more than simple geometry.
I will give you a couple of hints:
1. You will be suprised at how elegant the proof is.
2. Try finding the areas of the smaller triangles within ABC.
3. There are most likely more than one way to find the solution. I however, only found one.
4. The method I used does not involve trig. (There might be methods which use trig that I am not aware of)
EDIT: Seeing desperate faces around anandtech,I have put up the solution!
This solution, like I promised, requires only pre-algebra/algebra one level math.
Note: this is probably not the only solution. If you know of another one, please PM me.
We first create a simple and very straightforward lemma:
1. If two triangles have the same length base and same length height, the areas are the same. Similarly, If two triangles have the same base, but one triangle's base is n times longer than the other triangle's, then the area of that triangle is n times the area of the other triangle. This is also true with heights.
Visulization 1
Visulization 2
Now, this lemma comes extremely handy for this problem.
It shows that:
(1). Area of CPE=Area of CPB
(2). Area of APB=Area of APE
(3). Area of BPC=3*Area of BPF
(4). Area of APC=3*Area of APF
Using (1) and (3), we can show
(5). Area of CPE=3*BPF
(The next part might be a bit confusing, refer to Visulization for help.)
If we let A=Area of BPF, then:
Area of BPF=A
Area of CPB=3A
Area of EPC=3A
(2) and (4) show us:
Area of BPF+Area of APF=Area of APE
and
Area of CPE+Area of APE=3*Area of APF.
Solving for the Area of FPA, we get:
Area of APE=3A
Area of FPA=2A
Therefore, we can show the area of the entire triangle is
Area of BPF+Area of FPA + area of APE + area of EPC + area of CPB
=3A+3A+3A+2A+A
=12A, where A is the area of BPF.
Note that Quadlaterial BPCA=9x, and Triangle BPC=3x.
Using the height part of our lemma (Visulization 2), PA/DP must be 3.
Therefore, PA=15, and PD=5.
Now comes the interesting part.
Take a look at the triangle APC. We know AP=15, PE=6, and PC=9.
Draw a median from point E to the midpoint of PC, which we call point R.
Visulization
Note PR=4.5, RC=4.5, and ER=15/2=7.5.
Do you notice anything?
Try harder.... what do you see?
Yes, that's right. 4.5-6-7.5 is a multiple of a pythogorean triplet, so therefore Angle RPE is a right angle, and Triangle RPE is a right triangle. Visulization
Now, everything becomes trival. Area of RPE is 1/2*4.5*6=27/2, so Area of CPE is 27. Since Area of ABC=12A, and Area of CPE=3A,
ABC=4*27=108.
QED
problem1.jpg
Find the area of the large triangle, in exact terms.
NOTE: The figure is not drawn to scale.
Give details on how you arrived at the answer. If you just post a number, I will not reply to your post, even if it's correct.
This question is solvable. It took me more than half a hour. It's a hard one though. However, it requires nothing more than simple geometry.
I will give you a couple of hints:
1. You will be suprised at how elegant the proof is.
2. Try finding the areas of the smaller triangles within ABC.
3. There are most likely more than one way to find the solution. I however, only found one.
4. The method I used does not involve trig. (There might be methods which use trig that I am not aware of)
EDIT: Seeing desperate faces around anandtech,I have put up the solution!
This solution, like I promised, requires only pre-algebra/algebra one level math.
Note: this is probably not the only solution. If you know of another one, please PM me.
We first create a simple and very straightforward lemma:
1. If two triangles have the same length base and same length height, the areas are the same. Similarly, If two triangles have the same base, but one triangle's base is n times longer than the other triangle's, then the area of that triangle is n times the area of the other triangle. This is also true with heights.
Visulization 1
Visulization 2
Now, this lemma comes extremely handy for this problem.
It shows that:
(1). Area of CPE=Area of CPB
(2). Area of APB=Area of APE
(3). Area of BPC=3*Area of BPF
(4). Area of APC=3*Area of APF
Using (1) and (3), we can show
(5). Area of CPE=3*BPF
(The next part might be a bit confusing, refer to Visulization for help.)
If we let A=Area of BPF, then:
Area of BPF=A
Area of CPB=3A
Area of EPC=3A
(2) and (4) show us:
Area of BPF+Area of APF=Area of APE
and
Area of CPE+Area of APE=3*Area of APF.
Solving for the Area of FPA, we get:
Area of APE=3A
Area of FPA=2A
Therefore, we can show the area of the entire triangle is
Area of BPF+Area of FPA + area of APE + area of EPC + area of CPB
=3A+3A+3A+2A+A
=12A, where A is the area of BPF.
Note that Quadlaterial BPCA=9x, and Triangle BPC=3x.
Using the height part of our lemma (Visulization 2), PA/DP must be 3.
Therefore, PA=15, and PD=5.
Now comes the interesting part.
Take a look at the triangle APC. We know AP=15, PE=6, and PC=9.
Draw a median from point E to the midpoint of PC, which we call point R.
Visulization
Note PR=4.5, RC=4.5, and ER=15/2=7.5.
Do you notice anything?
Try harder.... what do you see?
Yes, that's right. 4.5-6-7.5 is a multiple of a pythogorean triplet, so therefore Angle RPE is a right angle, and Triangle RPE is a right triangle. Visulization
Now, everything becomes trival. Area of RPE is 1/2*4.5*6=27/2, so Area of CPE is 27. Since Area of ABC=12A, and Area of CPE=3A,
ABC=4*27=108.
QED
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