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Difference between revisions of "1952 AHSME Problems/Problem 49"

(→‎Alternative but very similar Solution)
(→‎Alternative but very similar Solution)
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== Alternative but very similar Solution ==
 
== Alternative but very similar Solution ==
−
Let <math>[ABC]=K.</math> Then <math>[ADC] = \frac{1}{3}K,</math> and hence <math>[N_1DC] = \frac{1}{7} [ADC] = \frac{1}{21}K.</math> Similarly, <math>[N_2EA]=[N_3FB]=[N_1DC] = \frac{1}{21}K.</math> <math>[ADC]+[BEA]+[CFB]=\frac{1}{3}K+\frac{1}{3}K+\frac{1}{3}K = K.</math> Then we can implement a similar but different area addition postulate to the first solution. It will be <math>[ADC]+[BEA]+[CFB]=K-\frac{1}{21}K-\frac{1}{21}K-\frac{1}{21}K+[N_1N_2N_3].</math> Using transitive property <math>K-\frac{1}{21}K-\frac{1}{21}K-\frac{1}{21}K+[N_1N_2N_3] = K.</math> Subtracting and adding on both sides gives: <cmath>[N_1N_2N_3] = \boxed{\textbf{(C) }\frac{1}{7}[ABC]}</cmath> ~many credits to the first solution ~Lopkiloinm
+
Let <math>[ABC]=K.</math> Then <math>[ADC] = \frac{1}{3}K,</math> and hence <math>[N_1DC] = \frac{1}{7} [ADC] = \frac{1}{21}K.</math> Similarly, <math>[N_2EA]=[N_3FB]=[N_1DC] = \frac{1}{21}K.</math> <math>[ADC]+[BEA]+[CFB]=\frac{1}{3}K+\frac{1}{3}K+\frac{1}{3}K = K.</math> Then we can implement a similar but different area addition postulate to the first solution. It will be <math>[ADC]+[BEA]+[CFB]=K-\frac{1}{21}K-\frac{1}{21}K-\frac{1}{21}K+[N_1N_2N_3]</math> (PIE in action). Using transitive property <math>K-\frac{1}{21}K-\frac{1}{21}K-\frac{1}{21}K+[N_1N_2N_3] = K.</math> Subtracting and adding on both sides gives: <cmath>[N_1N_2N_3] = \boxed{\textbf{(C) }\frac{1}{7}[ABC]}</cmath> ~many credits to the first solution ~Lopkiloinm
  
 
== Solution 2 (best solution)==
 
== Solution 2 (best solution)==

Revision as of 17:22, 6 November 2020

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, \[[N_1N_2N_3] = K - 3\left(\frac{1}{21}K\right) - 3\left(\frac{5}{21}\right)K = K - \frac{6}{7}K = \boxed{\textbf{(C) }\frac{1}{7}[ABC]}\]

Alternative but very similar Solution

Let $[ABC]=K.$ Then $[ADC] = \frac{1}{3}K,$ and hence $[N_1DC] = \frac{1}{7} [ADC] = \frac{1}{21}K.$ Similarly, $[N_2EA]=[N_3FB]=[N_1DC] = \frac{1}{21}K.$ $[ADC]+[BEA]+[CFB]=\frac{1}{3}K+\frac{1}{3}K+\frac{1}{3}K = K.$ Then we can implement a similar but different area addition postulate to the first solution. It will be $[ADC]+[BEA]+[CFB]=K-\frac{1}{21}K-\frac{1}{21}K-\frac{1}{21}K+[N_1N_2N_3]$ (PIE in action). Using transitive property $K-\frac{1}{21}K-\frac{1}{21}K-\frac{1}{21}K+[N_1N_2N_3] = K.$ Subtracting and adding on both sides gives: \[[N_1N_2N_3] = \boxed{\textbf{(C) }\frac{1}{7}[ABC]}\] ~many credits to the first solution ~Lopkiloinm

Solution 2 (best solution)

We can force this triangle to be equilateral because the ratios are always $3:3:1$ no matter which rotation, and with the symmetry of the equilateral triangle, it is safe to assume that this may be the truth (symmetry is very important, kids). 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: ~Lopkiloinm \[\boxed{\textbf{(C) }\frac{1}{7}[ABC]}\] (Note: the presence of $7$ in the denominator gives hints on the answer, so this solution still seems good)

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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