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| − | ==Problem==
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| − | In a magic triangle, each of the six [[whole number|whole numbers]] <math>10-15</math> is placed in one of the [[circle|circles]] so that the sum, <math>S</math>, of the three numbers on each side of the [[triangle]] is the same. The largest possible value for <math>S</math> is
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| − | <asy>
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| − | draw(circle((0,0),1));
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| − | draw(dir(60)--6*dir(60));
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| − | draw(circle(7*dir(60),1));
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| − | draw(8*dir(60)--13*dir(60));
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| − | draw(circle(14*dir(60),1));
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| − | draw((1,0)--(6,0));
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| − | draw(circle((7,0),1));
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| − | draw((8,0)--(13,0));
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| − | draw(circle((14,0),1));
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| − | draw(circle((10.5,6.0621778264910705273460621952706),1));
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| − | draw((13.5,0.86602540378443864676372317075294)--(11,5.1961524227066318805823390245176));
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| − | draw((10,6.9282032302755091741097853660235)--(7.5,11.258330249197702407928401219788));
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| − | </asy>
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| − | <math>\text{(A)}\ 36 \qquad \text{(B)}\ 37 \qquad \text{(C)}\ 38 \qquad \text{(D)}\ 39 \qquad \text{(E)}\ 40</math>
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| − | ==Solution 1==
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| − | Let the number in the top circle be <math>a</math> and then <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math>, and <math>f</math>, going in clockwise order. Then, we have <cmath>S=a+b+c</cmath> <cmath>S=c+d+e</cmath> <cmath>S=e+f+a</cmath>
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| − | Adding these [[equation|equations]] together, we get
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| − | <cmath>\begin{align*}
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| − | 3S &= (a+b+c+d+e+f)+(a+c+e) \\
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| − | &= 75+(a+c+e) \\
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| − | \end{align*}</cmath>
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| − | where the last step comes from the fact that since <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math>, and <math>f</math> are the numbers <math>10-15</math> in some order, their [[sum]] is <math>10+11+12+13+14+15=75</math>
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| − | The left hand side is [[divisible]] by <math>3</math> and <math>75</math> is divisible by <math>3</math>, so <math>a+c+e</math> must be divisible by <math>3</math>. The largest possible value of <math>a+c+e</math> is then <math>15+14+13=42</math>, and the corresponding value of <math>S</math> is <math>\frac{75+42}{3}=39</math>, which is choice <math>\boxed{\text{D}}</math>.
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| − | It turns out this sum is attainable if you let <cmath>a=15</cmath> <cmath>b=10</cmath> <cmath>c=14</cmath> <cmath>d=12</cmath> <cmath>e=13</cmath> <cmath>f=11</cmath>
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| − | ==Solution 2==
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| − | To make the sum the greatest, put the three largest numbers <math>(13,14</math> and <math>15)</math> in the corners. Then, balance the sides by putting the least integer <math>(10)</math> between the greatest sum <math>(14</math> and <math>15)</math>. Then put the next least integer <math>(11)</math> between the next greatest sum (<math>13 +15</math>). Fill in the last integer <math>(12)</math> and you can see that the sum of any three numbers on a side is (for example) <math>14 +10 + 15 = 39</math>
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| − | <math>\boxed{\text{D}}</math>.
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| − | -by goldenn
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| − | ==See Also==
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| − | {{AJHSME box|year=1985|num-b=23|num-a=25}}
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| − | [[Category:Introductory Number Theory Problems]]
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| − | [[Category:Introductory Algebra Problems]]
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| − | {{MAA Notice}}
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