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1952 AHSME Problems/Problem 49

Revision as of 15:59, 6 November 2020 by Lopkiloinm (talk | contribs) (→‎Solution)

Problem

[asy] unitsize(27); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A,B,C,D,E,F,X,Y,Z; A=(3,3); B=(0,0); C=(6,0); D=(4,0); E=(4,2); F=(1,1); draw(A--B--C--cycle); draw(A--D); draw(B--E); draw(C--F); X=intersectionpoint(A--D,C--F); Y=intersectionpoint(B--E,A--D); Z=intersectionpoint(B--E,C--F); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,S); label("$E$",E,NE); label("$F$",F,NW); label("$N_1$",X,NE); label("$N_2$",Y,WNW); label("$N_3$",Z,S); [/asy]

In the figure, $\overline{CD}$, $\overline{AE}$ and $\overline{BF}$ are one-third of their respective sides. It follows that $\overline{AN_2}: \overline{N_2N_1}: \overline{N_1D} = 3: 3: 1$, and similarly for lines BE and CF. Then the area of triangle $N_1N_2N_3$ is:

$\text{(A) } \frac {1}{10} \triangle ABC \qquad \text{(B) } \frac {1}{9} \triangle ABC \qquad \text{(C) } \frac{1}{7}\triangle ABC\qquad \text{(D) } \frac{1}{6}\triangle ABC\qquad \text{(E) } \text{none of these}$

Solution

Let $[ABC]=K.$ Then $[ADC] = \frac{1}{3}K,$ and hence $[N_1DC] = \frac{1}{7} [ADC] = \frac{1}{21}K.$ Similarly, $[N_2EA]=[N_3FB] = \frac{1}{21}K.$ Then $[N_2N_1CE] = [ADC] - [N_1DC]-[N_2EA] = \frac{5}{21}K,$ and same for the other quadrilaterals. Then $[N_1N_2N_3]$ is just $[ABC]$ minus all the other regions we just computed. That is, \[[N_1N_2N_3] = K - 3\left(\frac{1}{21}K\right) - 3\left(\frac{5}{21}\right)K = K - \frac{6}{7}K = \boxed{\textbf{(C) }\frac{1}{7}[ABC]}\]

Solution 2

We can force this triangle to be equilateral because the ratios are always $3:3:1$, and with the symmetry of the equilateral triangle, it is safe to assume that this may be the truth. Then, we can do a simple coordinate bash. Let $B$ be at $(0,0)$, $A$ be at $(3,3\sqrt{3})$, and $C$ be at $(6,0)$. We then create a new point $O$ at the center of everything. It should be noted because of similarity between $\triangle N_{1}N_{2}N_{3}$ and $\triangle ABC$, we can find the scale factor between the two triangle by simply dividing $AO$ by $N_{2}O$ (nitrous oxide). First, we need to find the coordinates of $O$ and $N_{2}$. $O$ is easily found at $(3,\sqrt{3})$ and $N_{2}$ be found by calculating equation of $BE$ and $AD$.$E$ is located $(4,2\sqrt{3})$ so $BE$ is $y=\frac{x\sqrt{3}}{2}$. $D$ be at $(4,0)$ and the slope is $-3\sqrt{3}$. We see that they be at the same $x$-value. Quick maths calculate the x value to be $4-\frac{2\sqrt{3}}{3\sqrt{3}+\frac{\sqrt{3}}{2}}$ which be $3\frac{3}{7}$. Another quick maths caculation of the $y$-value lead it be equal $2\sqrt{3}*\frac{3\sqrt{3}}{\frac{\sqrt{3}}{2}+3\sqrt{3}}$ which be $1\frac{5}{7}\sqrt{3}$. Peferct, so now $N_{2}$ be at $(3\frac{3}{7},1\frac{5}{7}\sqrt{3})$. Subtracting the coordinate with the center give you $(\frac{3}{7}, \frac{5}{7}\sqrt{3})$. I don't even want to do this anymore so here is the answer: \[\boxed{\textbf{(C) }\frac{1}{7}[ABC]}\]

See also

1952 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 48 		Followed by
Problem 50
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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