Difference between revisions of "1985 AJHSME Problem 24"

(Solution)
(Added solution)
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== Solution ==
 
== Solution ==
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(Incomplete solution)
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To maximize <math>S,</math> we must place the largest numbers a following equation:
 
To maximize <math>S,</math> we must place the largest numbers a following equation:
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== Solution 2 ==
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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.
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[[File:24 fig1.png]]
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Then, we can label the left empty circle as <math>a</math> and the right empty circle as <math>a-1,</math> 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 <math>a+1</math> because <math>13+14=15+13-1.</math> We only can use the numbers <math>10, 11,</math> and <math>12</math> now, so <math>a=11,</math> <math>a+1=12,</math> and <math>a-1=10.</math> Therefore, the largest value for <math>S</math> is <math>15+11+13=10\cdot3+(5+3+1)=\boxed{39~\textbf{(D)}}.</math>
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But, we have to make sure that all three sides sum to 39. The completed triangle is shown in Figure 2.
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[[File:24 fig2.png]]

Revision as of 21:48, 26 July 2021

Problem

In a magic triangle, each of the six whole numbers $10-15$ is placed in one of the circles so that the sum, $S$, of the three numbers on each side of the triangle is the same. The largest possible value for $S$ 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]

$\text{(A)}\ 36 \qquad \text{(B)}\ 37 \qquad \text{(C)}\ 38 \qquad \text{(D)}\ 39 \qquad \text{(E)}\ 40$

Solution

(Incomplete solution)

To maximize $S,$ we must place the largest numbers a following equation:

Solution 2

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.

24 fig1.png

Then, we can label the left empty circle as $a$ and the right empty circle as $a-1,$ 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 $a+1$ because $13+14=15+13-1.$ We only can use the numbers $10, 11,$ and $12$ now, so $a=11,$ $a+1=12,$ and $a-1=10.$ Therefore, the largest value for $S$ is $15+11+13=10\cdot3+(5+3+1)=\boxed{39~\textbf{(D)}}.$

But, we have to make sure that all three sides sum to 39. The completed triangle is shown in Figure 2.

24 fig2.png