![]() Revision 3 - National Standards for Traffic Control Devices the Manual on. Xinding Sun from University of California, Santa Barbara found yet another twist to the problem.įurther variants are discussed at the CTK Exchange - the old archive. On January 4, 2021, the FHWA issued Official Interpretation 2(09)-174 (I) to. Thought and points out even more of self-imposed restrictions that I missed considering the 9-point problem Now, try to think of a restriction you imposed on yourself which was not inherent to the problem.Īs an aside, Lars Hellvig from Stockholm, Sweden picks up this line of The solution lies in the observation that it's permissible to cross square boundaries. ![]() For one always comes up with 5 lines instead of 4. The problem is to connect the dots with no more than 4 straight lines without lifting your hand from the paper. One classical example is where nine dots are arranged on the sides and the center of a square as in the picture below. R2 Admission TipsThe best hint, in my view, is to mention that often, when solving a problem, we implicitly impose constraints that have not come with the problem. Wednesday, Nov 22,ġ0:30am NY 3:30pm London 9pm Mumbai ✅ Subscribe to us on YouTube AND Get FREE Access to Premium GMAT Question Bank for 7 Days ✅ Former Admissions Dean provides important application tips for Round 2 MBA applications. Symmetrically, the same logic will apply as applied in case 3.Īnother 6 invalid cases that must be removed. Any 3 dots chosen from these 4 dots that lie on a diagonal straight line will NOT form a triangle. The upward sloping diagonal of the 4 by 4 array contains 4 dots. Each will be an invalid case -> 2 invalid There are 2 Diagonally upward slopping lines that sandwich the Center Diagonal.Įach of those lines has only 3 dots. There will be another 16 invalid groupings.Ĭase 3: 3 dots that lie on Diagonally Upwards Sloping Lines Since there are 4 rows: (4 rows) * (4 combinations per row that do not form a triangle) = 16 invalid How many ways can we have unique groups of 3 dots chosen out of 4 in a row? Any 3 unique dots chosen from those 4 dots will lie on a straight line and will NOT form a valid triangle. Out of these 560 groupings, how many groupings involve 3 dots located on the same Lineįor any one vertical row, there are 4 dots. (16 c 3) = 16! / (3! 13!) = 560 unique combinations can be made of 3 dots (Unfavorable Outcomes in which the 3 dots chosen all lie on the same line) (total possible unique combinations of 3 dots chosen out of the 16 total) Given that there are 16 dots equally spaced in an array, any unique combination of E does chosen can be a valid triangle with positive area UNLESS those 3 points chosen are COLINEAR Number of ways to connect any 3 distinct dots in the figure into a triangle = 560 - 16 - 16 - 6 - 6 = 516 Number of ways to connect any 3 distinct dots into bottom-left to top-right lines (non-triangles) = 1+4C3+1 = 6 Number of ways to connect any 3 distinct dots into top-left to bottom-right lines (non-triangles) = 1+4C3+1 = 6 ![]() Number of ways to connect any 3 distinct dots into a vertical line (non-triangles) = 4C3 *4 = 16 Number of ways to connect any 3 distinct dots into a horizontal line (non-triangles) = 4C3*4 = 16 For a discussion of the uses & abuses of formulas on the GMAT Quant section, as well as the complete solution to this problem, see: ![]() ![]() Many GMAT math problems, such as this one, cannot be solved by formulas alone. (Notice that three points all on the same line cannot form a triangle in other words, a triangle must have some area.) How many triangles, of absolutely any shape, can be created from three dots in this diagram? Different orientations (reflections, rotations, etc.) and/or positions count as different triangles. In the above diagram, the 16 dots are in rows and columns, and are equally spaced in both the horizontal & vertical direction. ![]()
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