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Magoosh GMAT Instructor
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Re: The above 11 x 11 grid of dots is evenly spaced: each dot is [#permalink]
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well there are 10 * 9 possible starting places for the triangle.

There are 4 different orientations (IE rotate it 90 degrees four times).

BUt there are twice as many triangles in actuality because we need to invert/flip it.

So 10*9*4*2 = 720
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Re: The above 11 x 11 grid of dots is evenly spaced: each dot is [#permalink]
I did not understand the solution can you please help
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Re: The above 11 x 11 grid of dots is evenly spaced: each dot is [#permalink]
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jasu0072005 wrote:
I did not understand the solution can you please help

Please see the entire blog article:
https://magoosh.com/gmat/2013/how-to-do- ... th-faster/
You will find a full explanation there.
Mike :-)
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Re: The above 11 x 11 grid of dots is evenly spaced: each dot is [#permalink]
Hi Mike,

Suppose i want to go by another approach,one that does not take rectangles into the picture.

Let us make this our XY coordinate system.
Let us call the triangle ABC so that A has the right angle AB is parallel to the X axis and AC is parallel to the Y axis(in one of the possible orientations).

Case 1: horizontal is 2 units long and vertical is one unit long(for triangle ABC)

Point A: It has 9 options for its X coordinate and 10 options for the Y coordinate. So 90 possible options for A
Point B: will always be restricted by the length given in the question. So only 1 option possible.
Point C:will always be restricted by the vertical length given . so only one option.

So total number of triangles possible in this orientation=90.

SImilarly, we will have 90 triangles possible with this fixed orientation and with the vertical length =2units.

Summing, we have 90+90=180 triangles possible.
Now, this where I get stuck.

Could you help me how to go ahead with the number of different orientations possible?
Maybe with permutations perhaps??

Thanks in advance
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Re: The above 11 x 11 grid of dots is evenly spaced: each dot is [#permalink]
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Dear bhang,

You are not counting all orientations. For example, for Case I, 2 units hor & 1 units vert ----- Is B to the right or left of A? Is C above or below A? You see, for any Case I triangle,you are only counting B to the right of A and C above A, but either one of those could still be different.
Did you read my full solution on the linked page above?

Mike :-)


12bhang wrote:
Hi Mike,

Suppose i want to go by another approach,one that does not take rectangles into the picture.

Let us make this our XY coordinate system.
Let us call the triangle ABC so that A has the right angle AB is parallel to the X axis and AC is parallel to the Y axis(in one of the possible orientations).

Case 1: horizontal is 2 units long and vertical is one unit long(for triangle ABC)

Point A: It has 9 options for its X coordinate and 10 options for the Y coordinate. So 90 possible options for A
Point B: will always be restricted by the length given in the question. So only 1 option possible.
Point C:will always be restricted by the vertical length given . so only one option.

So total number of triangles possible in this orientation=90.

SImilarly, we will have 90 triangles possible with this fixed orientation and with the vertical length =2units.

Summing, we have 90+90=180 triangles possible.
Now, this where I get stuck.

Could you help me how to go ahead with the number of different orientations possible?
Maybe with permutations perhaps??

Thanks in advance
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Re: The above 11 x 11 grid of dots is evenly spaced: each dot is [#permalink]
mikemcgarry wrote:
Dear bhang,

You are not counting all orientations. For example, for Case I, 2 units hor & 1 units vert ----- Is B to the right or left of A? Is C above or below A? You see, for any Case I triangle,you are only counting B to the right of A and C above A, but either one of those could still be different.
Did you read my full solution on the linked page above?

Mike :-)


12bhang wrote:
Hi Mike,

Suppose i want to go by another approach,one that does not take rectangles into the picture.

Let us make this our XY coordinate system.
Let us call the triangle ABC so that A has the right angle AB is parallel to the X axis and AC is parallel to the Y axis(in one of the possible orientations).

Case 1: horizontal is 2 units long and vertical is one unit long(for triangle ABC)

Point A: It has 9 options for its X coordinate and 10 options for the Y coordinate. So 90 possible options for A
Point B: will always be restricted by the length given in the question. So only 1 option possible.
Point C:will always be restricted by the vertical length given . so only one option.

So total number of triangles possible in this orientation=90.

SImilarly, we will have 90 triangles possible with this fixed orientation and with the vertical length =2units.

Summing, we have 90+90=180 triangles possible.
Now, this where I get stuck.

Could you help me how to go ahead with the number of different orientations possible?
Maybe with permutations perhaps??

Thanks in advance


Hey mike why didnt you use the number of points (121) and used only orientations and inversion?
For eg in case -2<x<4 and 6<y<16, we count all the points?
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The above 11 x 11 grid of dots is evenly spaced: each dot is [#permalink]
My take on this one.. can be solved in the 90 sec time...
Congruent triangles will fit on each other perfectly essentially triangles of same dimensions.
the only orientations possible are picking 2 spaces i.e.,3 dots horizontally or vertically.
If you see diagonally the length of the sides will be longer. hence only two possible orientations.
Horizontally
out of 10 spaces we need 2 continuous spaces...so 9c2(we need two continuous spaces hence 2 spaces will be considered as 1)
the one space 10c1
Total :9c2 * 10C1 =360
Vertically same :360
total 360*2=720

*Please feel free to point out any errors in my reasoning*
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Re: The above 11 x 11 grid of dots is evenly spaced: each dot is [#permalink]
@
goforgmat wrote:
My take on this one.. can be solved in the 90 sec time...
Congruent triangles will fit on each other perfectly essentially triangles of same dimensions.
the only orientations possible are picking 2 spaces i.e.,3 dots horizontally or vertically.
If you see diagonally the length of the sides will be longer. hence only two possible orientations.
Horizontally
out of 10 spaces we need 2 continuous spaces...so 9c2(we need two continuous spaces hence 2 spaces will be considered as 1)
the one space 10c1
Total :9c2 * 10C1 =360
Vertically same :360
total 360*2=720

*Please feel free to point out any errors in my reasoning*


Can you please explain why we need 10 spaces horizontally? (or vertically)? can you please walk me through the way of thinking??
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Re: The above 11 x 11 grid of dots is evenly spaced: each dot is [#permalink]
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georgekaterji wrote:
@
goforgmat wrote:
My take on this one.. can be solved in the 90 sec time...
Congruent triangles will fit on each other perfectly essentially triangles of same dimensions.
the only orientations possible are picking 2 spaces i.e.,3 dots horizontally or vertically.
If you see diagonally the length of the sides will be longer. hence only two possible orientations.
Horizontally
out of 10 spaces we need 2 continuous spaces...so 9c2(we need two continuous spaces hence 2 spaces will be considered as 1)
the one space 10c1
Total :9c2 * 10C1 =360
Vertically same :360
total 360*2=720

*Please feel free to point out any errors in my reasoning*


Can you please explain why we need 10 spaces horizontally? (or vertically)? can you please walk me through the way of thinking??

We need to connect three dots (horizontally or vertically), three dots continuously has 2 spaces continuously. We cant select any two spaces we need to select two continuous ones so we need to treat 2 continuous as 1 hence selecting 9c2.
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Re: The above 11 x 11 grid of dots is evenly spaced: each dot is [#permalink]
I got a dumb method but it works.

Forget about points, let's count squares.

Start with 1 square, you get 0 triangles.
Now, add a square beside it, then you are able to draw 4 triangles using two diagonals. (+4)
Then draw another square below our first square, OR beside our second square, we get 8 triangles. (+4)
Then, draw another square that connects to the existing horizontal row or the vertical column. You get 12 triangles (+4)
If you keep this going, you get a vertically inverted L shape.
you can also notice a pattern where with every additional square you get 4 more triangles.

So, according to the grid, to the right of the square 1, we have 9 squares. Below square 1, we also have 9 squares.
A total of 18 additional squares will get you 18*4=72 triangles.

Now that we have the L shaped frame, lets complete the inner part of the grid.
If you draw a square below the second square of the horizontal row OR to the right of the second square of the vertical column, you get 8 more triangles by connecting diagonals with the square above it, or the square beside it.
Then, draw another square beside OR below that, and you would get 8 more triangles.
You'll eventually realize that when a new square is drawn with two of its sides connected to the existing squares, you'd get to create 8 diagonals, instead of 4 from the "L" frame.

So, for the 11x11 grid, we have a total of 100 squares. Subtract the 19 squares we have in the inverted "L" shape, we have 81 squares left.
Since every new square within the inverted "L" frame produces 8 new triangles, we get 81*8=648 triangles.

Total=72+648=720 triangles.

Answer, D.
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