Difference between revisions of "2020 AMC 8 Problems/Problem 16"

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==Solution 1==
 
==Solution 1==
We notice that 3 lines pass through B, and 2 lines pass through all other points. In addition, we are given that <math>A+B+C+D+E+F=1+2+3+4+5+6=21</math>. This means that <cmath>2A+3B+2C+2D+2E+2F=47</cmath> <cmath>2(A+B+C+D+E+F)+B=47</cmath> <cmath>2(21)+B=47</cmath> <cmath>B=5</cmath>
+
We notice that 3 lines pass through B, and 2 lines pass through all other points. In addition, we are given that <math>A+B+C+D+E+F=1+2+3+4+5+6=21</math>. This means that <cmath>2A+3B+2C+2D+2E+2F=47</cmath> <cmath>2(A+B+C+D+E+F)+B=47</cmath> <cmath>2(21)+B=47</cmath> <cmath>B=5</cmath> <cmath>\boxed{\textbf{(E) }5}</cmath>
  
 
~samrocksnature
 
~samrocksnature

Revision as of 01:35, 18 November 2020

Each of the points $A,B,C,D,E,$ and $F$ in the figure below represents a different digit from $1$ to $6.$ Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is $47.$ What is the digit represented by $B?$ [asy] size(200); dotfactor = 10;  pair p1 = (-28,0); pair p2 = (-111,213); draw(p1--p2,linewidth(1));  pair p3 = (-160,0); pair p4 = (-244,213); draw(p3--p4,linewidth(1));  pair p5 = (-316,0); pair p6 = (-67,213); draw(p5--p6,linewidth(1));  pair p7 = (0, 68); pair p8 = (-350,10); draw(p7--p8,linewidth(1));  pair p9 = (0, 150); pair p10 = (-350, 62); draw(p9--p10,linewidth(1));  pair A = intersectionpoint(p1--p2, p5--p6); dot("$A$", A, 2*W);  pair B = intersectionpoint(p5--p6, p3--p4); dot("$B$", B, 2*WNW);  pair C = intersectionpoint(p7--p8, p5--p6); dot("$C$", C, 1.5*NW);  pair D = intersectionpoint(p3--p4, p7--p8); dot("$D$", D, 2*NNE);  pair EE = intersectionpoint(p1--p2, p7--p8); dot("$E$", EE, 2*NNE);  pair F = intersectionpoint(p1--p2, p9--p10); dot("$F$", F, 2*NNE); [/asy]

$\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5$

Solution 1

We notice that 3 lines pass through B, and 2 lines pass through all other points. In addition, we are given that $A+B+C+D+E+F=1+2+3+4+5+6=21$. This means that \[2A+3B+2C+2D+2E+2F=47\] \[2(A+B+C+D+E+F)+B=47\] \[2(21)+B=47\] \[B=5\] \[\boxed{\textbf{(E) }5}\]

~samrocksnature

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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