Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Alternative but very similar Solution)
(→‎Solution 2 (best solution))
Line 30: Line 30:
  
 
== Solution 2 (best solution)==
 
== Solution 2 (best solution)==
−
We can force this triangle to be equilateral because the ratios are always <math>3:3:1</math> 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 <math>B</math> be at <math>(0,0)</math>, <math>A</math> be at <math>(3,3\sqrt{3})</math>, and <math>C</math> be at <math>(6,0)</math>. We then create a new point <math>O</math> at the center of everything. It should be noted because of similarity between <math>\triangle N_{1}N_{2}N_{3}</math> and <math>\triangle ABC</math>, we can find the scale factor between the two triangle by simply dividing <math>AO</math> by <math>N_{2}O</math> (nitrous oxide). First, we need to find the coordinates of <math>O</math> and <math>N_{2}</math>. <math>O</math> is easily found at <math>(3,\sqrt{3})</math> and <math>N_{2}</math> be found by calculating equation of <math>BE</math> and <math>AD</math>.<math>E</math> is located <math>(4,2\sqrt{3})</math> so <math>BE</math> is <math>y=\frac{x\sqrt{3}}{2}</math>. <math>D</math> be at <math>(4,0)</math> and the slope is <math>-3\sqrt{3}</math>. We see that they be at the same <math>x</math>-value. Quick maths calculate the x value to be <math>4-\frac{2\sqrt{3}}{3\sqrt{3}+\frac{\sqrt{3}}{2}}</math> which be <math>3\frac{3}{7}</math>. Another quick maths caculation of the <math>y</math>-value lead it be equal <math>2\sqrt{3}*\frac{3\sqrt{3}}{\frac{\sqrt{3}}{2}+3\sqrt{3}}</math> which be <math>1\frac{5}{7}\sqrt{3}</math>. Peferct, so now <math>N_{2}</math> be at <math>(3\frac{3}{7},1\frac{5}{7}\sqrt{3})</math>. Subtracting the coordinate with the center give you <math>(\frac{3}{7}, \frac{5}{7}\sqrt{3})</math>. I don't even want to do this anymore so here is the answer: ~Lopkiloinm <cmath>\boxed{\textbf{(C) }\frac{1}{7}[ABC]}</cmath> (Note: the presence of <math>7</math> in the denominator gives hints on the answer, so this solution still seems good)
+
We can force this triangle to be equilateral because the ratios are always <math>3:3:1</math> 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 <math>B</math> be at <math>(0,0)</math>, <math>A</math> be at <math>(3,3\sqrt{3})</math>, and <math>C</math> be at <math>(6,0)</math>. We then create a new point <math>O</math> at the center of everything. It should be noted because of similarity between <math>\triangle N_{1}N_{2}N_{3}</math> and <math>\triangle ABC</math>, we can find the scale factor between the two triangle by simply dividing <math>N_{2}O</math>(not nitrous oxide) by <math>AO</math>. First, we need to find the coordinates of <math>O</math> and <math>N_{2}</math>. <math>O</math> is easily found at <math>(3,\sqrt{3})</math> and <math>N_{2}</math> be found by calculating equation of <math>BE</math> and <math>AD</math>.<math>E</math> is located <math>(4,2\sqrt{3})</math> so <math>BE</math> is <math>y=\frac{x\sqrt{3}}{2}</math>. <math>D</math> be at <math>(4,0)</math> and the slope is <math>-3\sqrt{3}</math>. We see that they be at the same <math>x</math>-value. Quick maths calculate the x value to be <math>4-\frac{2\sqrt{3}}{3\sqrt{3}+\frac{\sqrt{3}}{2}}</math> which be <math>3\frac{3}{7}</math>. Another quick maths caculation of the <math>y</math>-value lead it be equal <math>2\sqrt{3}*\frac{3\sqrt{3}}{\frac{\sqrt{3}}{2}+3\sqrt{3}}</math> which be <math>1\frac{5}{7}\sqrt{3}</math>. Peferct, so now <math>N_{2}</math> be at <math>(3\frac{3}{7},1\frac{5}{7}\sqrt{3})</math>. Subtracting the coordinate with the center give you <math>(\frac{3}{7}, \frac{5}{7}\sqrt{3})</math>. Pythagorean theorem on that gives <math>\frac{\sqrt{84}}{7}</math> which simplifies to <math>\frac{2\sqrt{3}}{\sqrt{7}}.</math> <math>A</math> is vertically above <math>O</math> by <math>2\sqrt{3}</math> units. The scale factor is thus <math>\frac{1}{\sqrt{7}} giving us an answer of:~Lopkiloinm <cmath>\boxed{\textbf{(C) }\frac{1}{7}[ABC]}</cmath> (Note: the presence of </math>7$ in the denominator gives hints on the answer, so when you see it, 1/7 looks like the obvious choice)
  
 
== See also ==
 
== See also ==

Revision as of 19:57, 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 $N_{2}O$(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 the coordinate with the center 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 <cmath>\boxed{\textbf{(C) }\frac{1}{7}[ABC]}</cmath> (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)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1952_AHSME_Problems/Problem_49&oldid=136818"