Filling In The Blanks Around A Triangle

Logic Level 4

Each of the six spaces is filled with a distinct positive integer $\le 25.$ For each of the four smaller triangles formed, if their vertices are filled with the integers $x < y < z,$ we must have $x+y-z = 5.$ Find the maximum possible value of the sum of the numbers in the circles.

41 39 51 47 55

1 solution

Eli Ross Staff
Feb 9, 2016

Let the largest element in the all-circles triangle be $m.$ It follows that the sum of the values in the circles must be $2m + 5,$ so we want to maximize $m.$

When considering the two adjacent squares to $m,$ note that they must each be bigger than $m$ ; otherwise, the square would need an equal value to the circle opposite it in the diagram, which is not possible since the integers must be distinct. Call those values for the squares $n_1 > m$ and $n_2 > m,$ and call the other two circle values $v_1$ and $v_2,$ corresponding to the squares they share a triangle with.

Now, we have three equations: $v_{1}+v_{2} -5=m$ , $m+v_1-n_1=5,$ and $m+v_2-n_2 = 5.$ Thus, $3m-5=n_1+n_2.$

Taking $(n_1,n_2) = (25,24)$ gives $3m-5 = 49,$ so $m=18.$ Then, $v_1,v_2=12,11,$ and we can fill the final square with $6.$ The largest possible sum is thus $2\cdot 18 +5=41.$

Thanks for such a beautiful problem! I replied it wrong nonetheless since even if I went right on the reasoning i made an imprecision in calculation and achieved that 41 is too great and an impossible case and yes , my way of solving this was a little bit different and i think yours is a little better anyways too....

A A - 5 years, 3 months ago

I'm glad you enjoyed the problem! Would you like to share your approach?

Eli Ross Staff - 5 years, 3 months ago

Oops , i'm sorry.... i was wrong even on that approach but since i made u curious my mistake was that I somehow took it as right that in the triangle which has just "circle vertices" z can't be bigger than 20...... from.. which i therefore deduced some other stuff so sorry again for wasting your time...

Actually as I just found out on the approach I made it can still be solved so I'll get to the point making before the remark that i refer to the triangle made from circles as "circle triangle" and the ones who have a square as "square triangles" . Now it can be firstly observed that the x (smallest) , y(medium) and z(biggest) numbers of the "circle triangle" can't be the x , y and z in the other triangles (since by the same line of reasoning you made if they are then we achieve a symmetry and we should put the same values we have for the x and y of the "circle triangle" in the squares having the same numbers and by this we wouldn't met the conditions of the problem) (1) and that since the largest possible number is 25 the largest possible sum made by the 2 smaller numbers in any triangle (including z) can't be greater than 30(2). Now by taking into account (1) and (2) this leads to interpreting the things like this : since z is not the greatest in the two square triangles in which it is it appears as the second smaller term in them and is then involved in the sums z + y and z + x of those "square triangles" which will have values less than 30 which combined with the fact that x + y is at a distance of 5 from z leads to the interpretation (and i consider this also the proper understanding of the equations) that z added with one of the numbers (say x) conditions the value of y and therefore by this conditions the result in the other "square triangle" (the sum z+ y for the choice of x as our reference value as it is in this example) which can't be greater than 30 also. This means , if we put the matter like this that x and y must be chosen so that for some choice of x the y isn't too large to give a bigger result than 30 in the other "squared triangle" which means that if we try to solve the problem by starting with a choice of z we have some guiding understanding for the way this values relate with each other in order to choose that value wisely. Now , having this understanding of the way the x , y and z of the "circle triangle" interrelate , we also know that for any choice of z we will have some value for the sum x+y from which the y will be at least (z+5)/2( if the number is even) and (z+5)/2(if the number is odd) ( and this is because the same sum , say S can be represented by the two numbers by a number of different pairs from which if one of the numbers is less than S/2 then by identity the other one must be greater with n units the other one is smaller) and from this number of pairs there must be therefore meaning that in the "square triangle" with the sum z+y will be at least z + (z+5)/2 which should be less than or equal with 30. So , in other words we are looking for the proper and maximum values of the pair (z , z+5/2) which actually is , as i find out pretty similar with your solution! although i arrived at this step at little bit differently by it's concreteness. And now we can say that z + (z+5)/2 = 30 (a simple equation somehow similar with yours) which means that z = 18 which by being a solution for the sum z +y=30 and 30 being the maximum value of that triangle leads to the rest of the solution by checking the relations....

So , i'm done now and indeed there is somehow a different solution although it's in it's way similar somehow.

The way our solutions differ to specify and to add as a remark.... is that while you approached it by the independently conceived relations between the "square triangles" coming to the 3 equations and then achieved a simple equation from them I took the relations between the numbers and tried to conceive their relations more in the terms of the values of x , y and z in the circle triangle.

Anyways , thanks for the problem!!

A A - 5 years, 3 months ago

Filling In The Blanks Around A Triangle

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