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1. Two lines \ell and m intersect at O at an angle of 28^\circ. Let A be a point inside the acute angle formed by \ell and m. Let B and C be the reflections of A in lines \ell and m, respectively. Find the number of degrees in \angle BAC.
2. A laser is shot from vertex A of square ABCD of side length 1, towards point P on \overline{BC} so that BP = 3/4. The laser reflects off the sides of the square, until it hits another vertex, at which point it stops. What is the length of the path the laser takes?
Thanks!
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Actually I've solved 1. The answer was 152.
Still not certain about no. 2, however.
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Hi;
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi SPARKS_CHAN;
For #2 I get 5, as per image:
What I don't know (and I've only done this graphically in Geogebra), is how to prove that the laser terminates at D.
Last edited by phrontister (2017-02-22 13:09:02)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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hi SPARKS_CHAN, and Phro,
There are lots of similar triangles here, so I dare say you could do this with lots of ratio calculations, and D will be a corner by symmetry.
If you want me to do the calcs, post again.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thanks, Bob. Yes, I can see that would work.
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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So, after sorting symmetries and ratios, I got:
Last edited by phrontister (2015-02-26 21:53:57)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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For #2, forget about the line being inside a square. Look at it as the straight line y=.75x, and ask: after (0,0), what is the first point on the line where x,y are both integers.
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Can you help me with this problem as well?
Equilateral triangle ABC has centroid G. Triangle A'B'C' is the image of triangle ABC upon a dilation with center G and scale factor -2/3. Let K be the area of the region that is within both triangles. Find K/[ABC].
I tried solving it in Geogebra but I think I screwed up somewhere.
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This has come up before here:
http://www.mathisfunforum.com/viewtopic.php?id=21327
Post 2 has the problem and post 7 has an outline of my solution.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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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.
If a second was a minute, and a minute was an hour, how many hours would be in one day? (There are multiple answers and they are debatable)
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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
If a second was a minute, and a minute was an hour, how many hours would be in one day? (There are multiple answers and they are debatable)
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hi ET ag
Welcome to the forum.
The forum LaTex doesn't support your draw commands so you won't get an image that way. Most members use Geogebra to construct their diagrams; save them to an on-line site; and then link from there using bcCode. A recent post has addressed this.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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