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A random triangle is drawn, and a blue square is drawn out from each of it's edges. Three more shaded triangles are formed by drawing the three lines connecting the vertices of the squares, and a red square is now drawn out from each of these lines too. The total area of the blue squares is A1, and the total area of the red squares is A2. Given that A2 = kA1, what is k?
To make it easier, it's one of these 5 {1, 1.5, 2, 3, holyshag}.
This is possible to do with coordinate geometry, setting the vertices of the triangle to (0, 0), (a, b), and (c, 0); however, the solution is very long and involves much painful arithmetic manipulation. I suspect there is a non-algebraic geometrical way to solve the problem, but I cannot think of a method, since geometry is one of my weaker areas. Any ideas?
maybe some math humour will inspire you, Kalkewl8ter!
Frank is presented a tough problem. He has three
watches:
A is broken
B goes off by one second everyday
C is always behind the correct time by one second
Frank doesnt know which one he should wear, so he took
it to The Computer. The Computer answered
immediately, concluding that A would be the best
choice. When asked why, it replied that A is correct
twice in a day, B is correct once every 120 years, and
C is never correct.
---
Laura is confused. She has a mother and a father. Her
parents each have two more parents whom she calls
grand-parents (2^2=4). Her grand-parents again have
two parents each (2^3=8). So using Mathematical
Induction, given nth generations ago, she would have
2^n parents. If every generation is approximately 30
years, then 600 years ago, she would have over one
million grand-parents (2^20=1,048,576)! At 500AD, or
50 generations ago, she would then have over one
trillion parants (2^50=1,125,899,906,842,624)?!?!
---
John likes to drink. Everytime he drinks, he feels
the need to drink one full bottle of beer. One day,
after he finishes one bottle, he lifts up the empty
bottle and says to himself "Half a bottle of beer,
viewed by pessimists as a half empty bottle of beer
and optimists as a half full bottle of beer.
Nonetheless, either pessmistic or optimistic, ...
1/2 empty bottle of beer = 1/2 full bottle of beer
Rules of Algebra says that if I multiply something on
B.S., the equality will still hold. So if I multiply
2...
empty bottle of beer = full bottle of beer
Therefore, the bottle I just drank isn't the same as
not drinking at all?!?!"
---
Bugz,
lol...Where did you get those? Or did you make them up? However, no matter what happens, I shall never be able to do Euclidean geometry!
i translated them from a chinese site. One of them, I have to give credit to Soroban, a retired math professor in US. I doubt that he made them up himself though.
Wow! That sounds interesting although I'm somewhat confused. Could you provide a drawing?Quote:
Originally posted by A$$Bandit
A random triangle is drawn...
Hey krtmrtz, aren't you busy?
BTW, I'm confused too, hopefully the drawing will sort that out ;-)
What, you want a diagram of a random triangle? Here are your instructions:
1) Draw a line of any length you like on some paper.
2) Draw another non-parallel line from the end of the first line, also of any length, and also on the paper.
3) Draw another non-parallel (again) line from the end of the second line to the beginning of the first one.
There's your random triangle.
Go on, I bet no-one can do this...
har,har....
do you really think the random triangle was the problem? Do you understand what figures are to be drawn? I don't and I think krtxmrtz doesn't get it too. So please provide a drawing, to show what you talk about with :
Quote:
....and a blue square is drawn out from each of it's edges. Three more shaded triangles are formed by drawing the three lines connecting the vertices of the squares, and a red square is now drawn out from each of these lines...
Well it didn't come with a diagram, but I'll go ahead and spell it out for you in a diagram. They're not all exactly squares but it's just a rough sketch using a grid...
k = 2
Because...?
Solution:
The three internal blue squares contribute a side to two of
the internal (gray) triangles.
i.e. We'll let the blue squares be square1, square2 and square3.
We'll further let the sides for each square be s1, s2, and s3.
The area of the entire BLUE section, therefore, is the
addition of the areas of each individual BLUE square or:
Area(BLUE) = s1^2 + s2^2 + s3^2;
The area of the RED squares cannot be determined until we
know the length of the side which neighbors the internal
triangles. This side of the triangle is the hypotenuse of the
triangle and is determined by taking the square root of the
sum of the squares of the other two sides.
We already know the other two sides of each of the triangles.
Triangle 1's sides are the sides of square 1 and square 2,
triangle 2's sides are the sides of square 2 and square 3, and
triangle 3's sides are the sides of square 3 and square 1.
In order to find the third side of each of these triangles we,
simply use the pythagorean theorem:
Triangle1 side3 = sqrt(s1^2 + s2^2)
Triangle2 side3 = sqrt(s2^2 + s3^2)
Triangle3 side3 = sqrt(s3^2 + s1^2)
These are also the lengths of the sides for each of the external
(RED) squares as previously mentioned.
So to find the area of each of the squares, simply square the
length of the hypotenuse of each of the triangles:
Square1 area = (sqrt(s1^2 + s2^2))^2 = (s1^2 + s2^2)
Square2 area = (sqrt(s2^2 + s3^2))^2 = (s2^2 + s3^2)
Square3 area = (sqrt(s3^2 + s1^2))^2 = (s3^2 + s1^2)
Adding up each of these areas give the TOTAL area of the RED
portion of the diagram or
Area(RED) = 2*s1^2 + 2*s2^2 + 2*s3^2
We already determined the area of hte blue region whcih is:
Area(BLUE) = s1^2 + s2^2 + s3^2
We can see that the area of the red region is simply two times
that of that area of the blue region so k=2.
The problem with your solution is, Pythagoras is only valid for triangles with an 90-degree angle.
And you can easily see, the grey triangles don't have to have such an angle!
So keep on trying!
I think it'S "Holyshag"!
Look at the attached diagram to for the notation of A,B,C
The angle in the grey triangles that is adjacent to an angle of white triangle following can be observed:
(1) Angle grey triangle=180 - (angle white triangle)
To calculate the missing side of the grey triangle formed by A and B using the Law of Cosine (hopefully the correct translation from german) use:
(2) X= A*A +B*B -2*A*B*cos(angle AB grey triangle)
using (1)
(3) X=A*A +B*B +2*A*B*cos(angle AB white triangle)
Note: cos(x)= -cos(180-x)
Used for all 3 red boxes, the formula to calculated the whole area is:
(4) X=2*(A*A+B*B+C*C)+ 2*(A*C*cos(AC)+A*B*cos(AB)+C*B*cos(CB)
I don't think that the last part of this formula can be modified to fit a formula (Area blue Squares)= k*(Area red Squares)
3, because most of the other answers have already gone... oh and that's the ratio of the sums of the areas of the squares - it's a cosine thing.
If p, q and r are the sides of a triangle and w is the angle between p and q, then:
r2=p2+q2 - 2pq cos(w) [1]
Using this equality, the problem can be solved in the following fashion (refer to the updated drawing):
http://www.vbforums.com/attachment.p...chmentid=18910
Let
A = A1 + A2 + A3 = a2 + b2 + c2 [2]
the area of the inner squares and
A' = A'1 + A'2 + A'3 = (a')2 + (b')2 + (c')2 [2]
that of the outer squares.
For the calculation of (a')2 we need to know u':
u' = 2*Pi - Pi/2 - Pi/2 - u = Pi - u
Then:
cos(u') = cos(Pi - u) = -cos(u)
Now applying the equality above [1]:
a'2=b'2+c'2 - 2b'c' cos(u')
But b=b' and c=c'. Subtituting these and cos(u') by -cos(u) we get:
a'2=b2+c2 + 2bc cos(u) [3]
But, using [1] again:
2bc cos(u) = b2+c2 - a2
Putting this into [3]:
a'2 = b2+c2 + b2+c2 - a2 = 2(b2+c2) - a2
By symmetry, (b')2 and (c')2 are:
b'2 = 2(a2+c2) - b2
c'2 = 2(a2+b2) - c2
Finally, susbtituting into [2]:
A' = (a')2 + (b')2 + (c')2 = 2(b2+c2) - a2 + 2(a2+c2) - b2 + 2(a2+b2) - c2 = 3(a2+b2+c2) = 3A
So, indeed, k=3
OKay, I think he did it. But I was close, I think
It was a question in the UK Senior Mathematical Challenge 2001, the last question too. Given that you have 24 other questions to do, some equally as hard as that one, in 1 hour 30 minutes, do you think you could do it?
Presnetly, definte NO. But since my math classes were in the last century.................................................
That means less than 4 minutes per question... even if you know all the answers right away it takes quite a while to write them down on paper (and without making any mistakes -only Mozart could do that!) and do the complete derivation. But I guess there may be people out there who can do it...Quote:
Originally posted by A$$Bandit
It was a question in the UK Senior Mathematical Challenge 2001, the last question too. Given that you have 24 other questions to do, some equally as hard as that one, in 1 hour 30 minutes, do you think you could do it?
Well you have 25 questions, you start with 25 marks. 4 for each correct answer, -1 for each wrong answer, 0 otherwise. Those marks are discrete. Maximum marks obviously 125, i got 64 which is considered to be a respectable score for a 16-year-old. One of my friends got 80.
OK, I mistakingly thought you were required to put down the entire derivation... so it's rather a checkbox answer contest, either you remember the anser or you don't... In these circumstances yes, I believe you could attain a high mark, time is not so decisive.Quote:
Originally posted by myself
... it takes quite a while to write them down on paper...