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I like how this was a discussion about a certain proof, and then it somehow transformed about a discussion of a movie.
Wait...I still couldn't get an answer for the soccer ball problem, can you guy elaborate on your hints?
Thanks,
ET ag
Hi Krash,
Given what was said above and the similarities that Bob gave, we get MX as 6, and we know that XY is 3. This shows that MY = 9, thus PQ = 18.
Ok. Thanks, I don't need a proof for **. I am in a little busy right now, so I will post my question later.
Thanks,
ET ag
bob bundy wrote:Imagine rotating triangle TAC through 90 degrees, anticlockwise, around point A.
T rotates to B and C rotates to V, so one triangle becomes the other.
So TC must also rotate 90 to become BV, hence they are perpendicular.
Bob
How do we know that you have to rotate it 90 degrees?
Well, if we wanted to rotate point
anti-clockwise about point , it would end up at point This is because is thus must "travel" that angle to reach point .(Sorry if the LaTeX doesn't seem very neat, I tried to make it "proper" using the correct commands on this database. Quite ironic how LaTeX is supposed to make math look neater when in fact it makes it look quite messy on this website, if you ask me).
(Funny how I have edited this message like 20 times and I have only sat at my desktop for about eight minutes).
From,
ET ag
Hi,
multiply the slopes of OA and BC and set it to -1
What exactly does this mean?
Also, from what the rest of thickhead's method is, how exactly should I find the coordinates of (h, k). Sorry, I just didn't really understand what you wrote.
Lastly,
I called the points (a,1/a); (b,1/b); and (c,1/c) and worked out the equation of the line perpendicular to AB through C. Once you have the form for that line, you can write down a similar equation for the line perpendicular to BC through A.
How did you work the equation for the line perpendicular to Ab through C?
Thanks,
ET ag
P.S. I only need a specification, since I see the method that you guys used, which was quite smart.
Hi,
I need a proof for the following problem before the midnight of Mar 3. I don't need a complete answer, though it would be helpful, since I need to write this in my own words. I need help PROVING, not knowing. So try to make it rigorous so that a person that has enough math knowledge to understand the problem and knows coordinate geometry can understand. However, don't "imply" anything or write it so that only a person with a PhD in Mattematics can understand.
Here's the question to prove: Let $A$, $B$, and $C$ be three points on the curve $xy = 1$ (which is a hyperbola). Prove that the orthocenter of triangle $ABC$ also lies on the curve $xy = 1$.
My feedback on this question: I know what an orthocenter is: where the three altitudes of a triangle meet, and it makes sense that it lies on a hyperbola. (Since I tried this on a piece of paper). But I can't just say, "If you try it, you'll see that this is possible." I must PROVE that this is possible and that is what I need help with. I explained above the deadline and the requirements, guidelines, and suggestions.
Thanks,
ET ag
P.S. I will check this thread every so often, so I should take too long to reply.
However, there is a rigorous method (that supports what wrlorimer said):
For each reflection of the laser off a side, we can reflect the square correspondingly, so that the reflected path becomes a straight line. For example, the path $APQR$ becomes line $APQ'R'$.
[asy]
unitsize(0.6 cm);
draw((0,0)--(8,6),red);
draw((4,3)--(8/3,4)--(0,2),red);
draw((0,0)--(8,0));
draw((0,4)--(8,4));
draw((4,8)--(8,8));
draw((0,0)--(0,4));
draw((4,0)--(4,8));
draw((8,0)--(8,8));
label("$A$", (0,0), SW);
label("$B$", (4,0), S);
label("$P$", (4,3), SE);
label("$Q$", (8/3,4), N);
label("$Q'$", (16/3,4), N);
label("$R$", (0,2), W);
label("$R'$", (8,6), E);
[/asy]
Looking at the straightened laser path, we see that it hits a corner only when it has traveled vertically and horizontally an integer distance. Since $AB = 1$ and $BP = 3/4$, after the path has travelled $k$ units horizontally, the path has travelled $3k/4$ units vertically. For both $k$ and $3k/4$ to be integers, $k$ must be a multiple of 4. In particular, $k$ has to be at least 4.
[asy]
unitsize(0.3 cm);
draw((0,0)--(16,12),red);
draw((4,3)--(8/3,4)--(0,2)--(8/3,0)--(4,1)--(0,4),red);
draw((0,0)--(16,0));
draw((0,4)--(16,4));
draw((0,8)--(16,8));
draw((0,12)--(16,12));
draw((0,0)--(0,12));
draw((4,0)--(4,12));
draw((8,0)--(8,12));
draw((12,0)--(12,12));
draw((16,0)--(16,12));
label("$4$", (8,0), S);
label("$3$", (16,6), E);
[/asy]
For $k = 4$, the path hits another corner that is 4 units away horizontally, and $3/4 \cdot 4 = 3$ units away vertically, so by Pythagoras, the length of the path is $\sqrt{3^2 + 4^2} = \boxed{5}$.
Sorry if my Asymptote or LaTeX didn't come out properly. Is there a way for it to come out properly and for my images to appear?
Thanks,
ET ag
Wrloimer's method is correct and easy, since the square has side length 1 and can go into any integer. However, if the side is $x$ then you have to find the first coordinates that divides $x.$
Sorry if the LaTeX didn't come out properly.
Hello mathstudent2000,
Since you use AoPS, you should be familiar with the AoPS Honor Code. Specifically, you should know that you should try to look up answers but rather try them. Remember, the teachers of your course are happy to help you on the Page Feed.
Also, the answer 0 is correct given Bob's reason.
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