Difference between revisions of "1985 AJHSME Problem 24"
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| − | Going clockwise around the circle and starting at the top, label the circles <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math>, and <math>f</math>. From the problem, we have that <math>a + b + c</math> = <math>c + d + e</math> = <math>e + f + a = S</math>, so <math>(a + b + c) + (c + d + e) + (e + f + a) = 2a + b + 2c + d + 2e + f = 3S</math>. The sum of <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math>, and <math>f</math> must be <math>75</math> (this is the sum of the possible values for the circles). From this, we have that <math>a + c + e + (a + b + c + d + e + f) = a + c + e + 75 = 3S</math>. To maximize <math>a + c + e</math>, we choose <math>a = 15</math>, <math>c = 14</math>, and <math>e = 13</math>. Then <math>(15 + 14 + 13) + 75 = 117 = 3S</math>, so <math>S = \boxed{39~\textbf{(D)}}</math>. | + | Going clockwise around the circle and starting at the top, label the circles <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math>, and <math>f</math>. From the problem, we have that <math>a + b + c</math> = <math>c + d + e</math> = <math>e + f + a = S</math>, so <math>(a + b + c) + (c + d + e) + (e + f + a) = 2a + b + 2c + d + 2e + f = 3S</math>. The sum of <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math>, and <math>f</math> must be <math>75</math> (this is the sum of the possible values for the circles). From this, we have that <math>a + c + e + (a + b + c + d + e + f) = a + c + e + 75 = 3S</math>. To maximize <math>a + c + e</math>, we choose <math>a = 15</math>, <math>c = 14</math>, and <math>e = 13</math> (this choice of assignment is arbitrary due to the commutativity of addition). Then <math>(15 + 14 + 13) + 75 = 117 = 3S</math>, so <math>S = \boxed{39~\textbf{(D)}}</math>. |
~ cxsmi | ~ cxsmi | ||
Revision as of 23:39, 17 January 2024
Problem
In a magic triangle, each of the six whole numbers 
is placed in one of the circles so that the sum, 
, of the three numbers on each side of the triangle is the same. The largest possible value for is
![[asy] draw(circle((0,0),1)); draw(dir(60)--6*dir(60)); draw(circle(7*dir(60),1)); draw(8*dir(60)--13*dir(60)); draw(circle(14*dir(60),1)); draw((1,0)--(6,0)); draw(circle((7,0),1)); draw((8,0)--(13,0)); draw(circle((14,0),1)); draw(circle((10.5,6.0621778264910705273460621952706),1)); draw((13.5,0.86602540378443864676372317075294)--(11,5.1961524227066318805823390245176)); draw((10,6.9282032302755091741097853660235)--(7.5,11.258330249197702407928401219788)); [/asy]](http://latex.artofproblemsolving.com/8/3/4/83415da2658f8c65d5efcd1a4cc144e572e5a2a0.png)
Solution 1
A numeral can appear in a maximum of 2 sides in this triangle, so we can put the largest 3 numbers, 15, 14, and 13 at the corners, as shown in Figure 1.
Figure 1
Then, we can label the left empty circle as 
and the right empty circle as 
because 14 is one more than 13, and we need to balance it out for the sum to be the same. The bottom empty circle can be named as 
because 
We only can use the numbers 
and 
now, so 
and 
Therefore, the largest value for is 
But, we have to make sure that all three sides sum to 39. The completed triangle is shown in Figure 2.
Figure 2
Solution 2
Going clockwise around the circle and starting at the top, label the circles , 
, 
, 
, 
, and 
. From the problem, we have that 
= 
= 
, so 
. The sum of , , , , , and must be 
(this is the sum of the possible values for the circles). From this, we have that 
. To maximize 
, we choose 
, 
, and 
(this choice of assignment is arbitrary due to the commutativity of addition). Then 
, so 
.
~ cxsmi
