In the grid of dots above, each dot has equal vertical &

Question

half-square grid of dots with triangle.JPG In the grid of dots above, each dot has equal vertical & horizontal spacing from the others. A small 45-45-90 triangle is drawn. Counting this triangle, how many triangles congruent to this one, of any orientation, can be constructed from dots in this grid? A. 112 B. 120 C. 240 D. 448 E. 480 For a discussion of difficult counting problems, as well as the solution to this problem, see: https://magoosh.com/gmat/2013/difficult- ... -problems/ Mike :-)

rockstar23 · Accepted Answer

half-square grid of dots with triangle.JPG Count the number of possible squares (just as the one highlighted above) in figure = 28 Each such square can contain 4 triangles = 28 x 4 = 112. On the diagonal of the half square we can have another 8 triangles. So total number of triangles = 112 + 8 = 120

mayer87 · Answer

well, sorry that i cant draw any pictures to illustrate but i will try my best to explain with word. 
Here is my approach: think of the distance between each dot as 1 unit, i try to image there is a square whose sides are 1 unit. When I try to count all those qualified squares. there are 28 of them. And in each such square, there are 4 ways to place the triangle under the 45-45-90 condition. So there are 28*4=112 of them, plus the left triangles which cannot form a square but also qualify, they are located at the right end from top to bottom, so we get another 8 of those triangles. In total 112+8=120

mvictor · Answer

good question...i almost started to count all possible 45-45-90 triangles that can be drawn..but since we need congruent...the task is easier...
we have 7, 6, 5, 4, 3, 2, 1 - squares that can be created using the dots. in which square, we can have 4 variations of the triangle.
sum of 7 = 7*8/2 = 7*4=28
now, 28*4 = 112. we can eliminate A right away, as we did not count the "outer" possible triangles.
we can draw 8 triangles with the "outer" points.
112+8=120.

B

p.s. is there a way to count all possible variations of 45-45-90 triangle??? I believe it should be possible using the similar method..but would take way too much to solve..

chetan2u · Answer

Hi,
 ONE way I can think of is - 
Look at the Hypotenuse...
the outermost has 9 dots and there on keeps reducing by 1 for each slant line below it..
the outermost cannot make any 45-45-90 triangles..
the second line has 8 dots and each dot can make ONE triangle with the outer most HYP= 8*1..
the Third line has 7 dots and each dot can make TWO triangle with the two hypotenuse outside= 7*2..
similarily 1 dots and that can make 8 triangles with 8 HYP= 1*8...
TOTAL =  8*1+7*2+6*3+5*4+4*5+3*6+2*7+1*8 = 2 ( 8*1+7*2+6*3+5*4 ) = 2(8+14+18+20) = 2*60 = 120

rohit8865 · Answer

a four dots will contribute to 4 congruent trangles as shown taking each set os dots as a new base 
secondly every extreme right side 3 dots will form only 1 triangle 
first row will make 1 triangle
second row will make 4+1=5 triangle
third row =9 triangles
a pattern here is every new row will add 4 new triangles.
there are 8 rows
total triangles=(1+5+9+....29)=120
Ans B

ashokjha1986 · Answer

Why bigger triangles are not considered in this? Does congruent means equal? I thought it to be similar triangle.

mikemcgarry · Answer

Dear ashokjha1986 , i'm happy to respond.

These are good terms to know, as a GMAT Quant problem might mention them: 1) congruent = same shape, same size (equal angles, equal lengths) 2) similar = same shape, different size (equal angles, proportional lengths) Does this make sense? Mike

Harshjha001 · Answer

from the 1st row we can get just 1 triangle . a=1 ( we will use it later)
from the 2nd row we get 5 triangles .
from the 3rd row we get 9 triangles and so on....

You can see there is a pattern -: for every consecutive row there is an increase of 4 triangles .
So total number of triangles = n/2 (2a + (n-1) d) -> formula for sum of an arithmetic sequence.
n=8 , a=1 , d=4 ; 4*( 2*1 + (8-1)*4) = 120.

IMO B

Fdambro294 · Answer

The easiest way for me to count the possibilities was a mixture of brute force and logic.

Ignoring the triangles that can be counted on the Slanted Edge and just focusing on each square:

Each triangle counted must be Congruent to the given triangle, which is (1/2) of a square that is 1 by 1

In each 1 by 1 unit square, there are 4 ways to draw the same congruent triangle: 2 ways using each diagonal.

So starting from the first row, there are 7 full, unique squares. (7 * 4)

The row next to it has 6 full, unique squares: (6 * 4)

This continues all the way to the very last 1 by 1 square on the far right.

Summing these possibilities up:
(7* 4) + (6 * 4) + (5 * 4) + (3 * 4) + (2 * 4) + (1 * 4) =

Take the 4 common:

4 * (7 + 6 + 5 + 4 + 3 + 2 + 1) =

4 * (28) = 112

Of course that’s listed as a trap answer lol

Lastly, you need to count the unique triangles that can be made on the Slanted Diagonal Edge.

There are 8 such triangles for a total of

(B) 120

Posted from my mobile device

ThatDudeKnows · Answer

This question is plenty straightforward to actually solve, so I probably wouldn't use it on this exact question, but I like taking opportunities to think like a test-writer and simply "beat" questions?

If you were going to create a trap answer or two for this question, one would probably be to include all the similar triangles in addition to the congruent ones, and a second would probably be to forget about the 8 triangles along the upper-right diagonal. Notice that there are only two answer choices that differ by 8? Those are probably the correct answer and the one that forgot about those 8 triangles. I'd have taken a bet that 120 was going to be the answer without doing any actual counting/math. Wouldntyaknow,...?

FWIW, a next step would be just to eliminate C, D, and E by eyeballing the number of squares to be roughly 25...DEFinitely fewer than the ~60 required to get to C and DEFINITELY fewer than the 100+ to get to D or E. We know we are down to A and B. Now I feel really confident that the +8 bit from above means that we should go with the answer that's 8 more than the other one.

KarishmaB · Answer

Another Method: Consider which all vertices can be the right angle vertex and how many right triangles can be placed on each. Screenshot 2022-06-22 at 9.45.44 AM.png The 14 dots inside the red rectangles can hold two right triangles each. This gives us 14 * 2 = 28 triangles The 21 dots inside the green triangle can hold 4 triangles each. This gives us 21 * 4 = 84 triangles The 7 dots inside the yellow oval cal hold 1 right triangle each. This gives us 7 triangles The dot on the bottom left corner can hold 1 right triangle. Total = 28 + 84 + 7 + 1 = 120

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