1952 AHSME Problems/Problem 49


[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:

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


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 $N_{2}O$(not nitrous oxide) by $AO$. 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 this coordinate with the coordinates of the center $O$ give you $(\frac{3}{7}, \frac{5}{7}\sqrt{3})$. Pythagorean theorem on that gives $\frac{\sqrt{84}}{7}$ which simplifies to $\frac{2\sqrt{3}}{\sqrt{7}}.$ $A$ is vertically above $O$ by $2\sqrt{3}$ units. The scale factor is thus $\frac{1}{\sqrt{7}}$ giving us an answer of: ~Lopkiloinm \[\boxed{\textbf{(C) }\frac{1}{7}[ABC]}\] (Note: the presence of $7$ in the denominator gives hints on the answer, so when you see it, 1/7 looks like the obvious choice) \[\] (Another note: the question gives you the ratio of $3:3:1$ so I did not need to use that many steps to calculate the coordinate of $N_2$, directly attaining it to be at $(3\frac{3}{7},1\frac{5}{7}\sqrt{3})$ After realizing this, this method should actually be pretty quick, may even be quicker than the non coordinate method. Regardless, if you don't get the ratio and you encounter a problem similar to this, the coordinate bash with all these calculations should get you the correct answer. Also, there is an even bashier way using all the points and shoelace formula. That takes too much time.)

Solution 3

Visual graphical proof. https://en.m.wikipedia.org/wiki/One-seventh_area_triangle#/media/File%3ATriangleOneSeventhAreaGraphicalSoln.png

Solution 4

Using Routh's Theorem

$x = \frac{BD}{CD} = \frac{\frac{2BC}{3}}{\frac{BC}{3}} = \frac{2}{1}$ and $x = y = z$


$[N_1N_2N_3] = [ABC]\cdot\frac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)}$

Substituting $x = y = z = 2$

$[N_1N_2N_3] = [ABC]\cdot\frac{7^2}{7^3}$ $[N_1N_2N_3] = [ABC]\cdot\frac{1}{7}$

Therefore the answer is,

(C) $[N_1N_2N_3] = \frac{[ABC]}{7}$

Solution 5 (Super fast)

We can use triangle area ratios. $[N_1N_2N_3]=[ABD]-[BFN_3]-[N_3N_2AF]-[N_3N_1DB]$.

So let $[ABC]=S$. Then $[ABD]=\frac{2}{3}S, [ADC]=\frac{1}{3}S, [N_1DC]=\frac{1}{21}S$.

Similarly, $[FBC]=\frac{1}{3}S$. So $[FBDN_1]=\frac{6}{21}S$. Similarly, we can use this argument to find $[ABE]=\frac{1}{3}S, [FBN_3]=\frac{1}{21}S, [AN_2E]=\frac{1}{21}S$. So $[FN_3N_2A]=\frac{5}{21}S$. So $[N_1N_2N_3]=\frac{2}{3}S-\frac{5}{21}S-\frac{6}{21}S=\frac{1}{7}S$.

So therefore, since $[N_1N_2N_3]=\frac{1}{7}[ABC]$, select $\boxed{C}$.

You can also use a similar idea (Principle of Inclusion-Exclusion) to find $[N_1N_2N_3]$. Notice that the triangle we desire is the intersection set $\triangle{AFC}, \triangle{ABD}, \triangle{BEC}$. So we just PIE it out.


See also

1952 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 48
Followed by
Problem 50
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