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Difference between revisions of "1974 AHSME Problems/Problem 19"

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Therefore, <math> CN^2=x^2+1=(2-\sqrt{3})^2+1=8-4\sqrt{3} </math>. The area of an equilateral triangle with side length <math> x </math> is equal to <math> \frac{x^2\sqrt{3}}{4} </math>, so the area of <math> \triangle CMN </math> is <math> \frac{(8-4\sqrt{3})(\sqrt{3})}{4}=2\sqrt{3}-3, \boxed{\text{A}} </math>.
 
Therefore, <math> CN^2=x^2+1=(2-\sqrt{3})^2+1=8-4\sqrt{3} </math>. The area of an equilateral triangle with side length <math> x </math> is equal to <math> \frac{x^2\sqrt{3}}{4} </math>, so the area of <math> \triangle CMN </math> is <math> \frac{(8-4\sqrt{3})(\sqrt{3})}{4}=2\sqrt{3}-3, \boxed{\text{A}} </math>.
  
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==Solution 2 (Using ratios)==
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==Solution 2 (Visualization + Using Ratios)==
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We know there is only one way to fit an equilateral triangle into a square: one of its vertices is a corner of the square and the other two vertices fall on opposite sides (try to imagine it in your head). It must be symmetrical along a diagonal of the square.
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Thus <math>\triangle CNB \sim\triangle CDM</math> where both are <math>15^{\circ}-75^{\circ}-90^{\circ}</math> triangles since <math>\angle MCN=60^{\circ}</math>
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Using the <math>15^{\circ}-75^{\circ}-90^{\circ}</math> ratios (<math>2-\sqrt3:1:\sqrt6-\sqrt2</math>), we have <math>CN = \sqrt6 - \sqrt2</math>.
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Thus <math>[\triangle CMN] = \frac{\sqrt3}{4}(\sqrt6 - \sqrt2)^2 = \frac{\sqrt3}{4}(8-4\sqrt3) = 2\sqrt3-3 \Rightarrow \fbox{A}</math>.
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<math>\sim</math> Jonysun
  
 
==See Also==
 
==See Also==

Latest revision as of 08:14, 29 October 2024

Problem

In the adjoining figure $ABCD$ is a square and $CMN$ is an equilateral triangle. If the area of $ABCD$ is one square inch, then the area of $CMN$ in square inches is

[asy] draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((.82,0)--(1,1)--(0,.76)--cycle); label("A", (0,0), S); label("B", (1,0), S); label("C", (1,1), N); label("D", (0,1), N); label("M", (0,.76), W); label("N", (.82,0), S);[/asy]

$\mathrm{(A)\ } 2\sqrt{3}-3 \qquad \mathrm{(B) \ }1-\frac{\sqrt{3}}{3} \qquad \mathrm{(C) \ } \frac{\sqrt{3}}{4} \qquad \mathrm{(D) \ } \frac{\sqrt{2}}{3} \qquad \mathrm{(E) \ }4-2\sqrt{3}$

Solution

Let $BN=x$ so that $AN=1-x$. From the Pythagorean Theorem on $\triangle NBC$, we get $CN=\sqrt{x^2+1}$, and from the Pythagorean Theorem on $\triangle AMN$, we get $MN=(1-x)\sqrt{2}$. Since $\triangle CMN$ is equilateral, we must have $\sqrt{x^2+1}=(1-x)\sqrt{2}\implies x^2+1=2x^2-4x+2\implies x^2-4x+1=0$. From the Pythagorean Theorem, we get $x=2-\sqrt{3}$, since we want the root that's less than $1$.

Therefore, $CN^2=x^2+1=(2-\sqrt{3})^2+1=8-4\sqrt{3}$. The area of an equilateral triangle with side length $x$ is equal to $\frac{x^2\sqrt{3}}{4}$, so the area of $\triangle CMN$ is $\frac{(8-4\sqrt{3})(\sqrt{3})}{4}=2\sqrt{3}-3, \boxed{\text{A}}$.

Solution 2 (Visualization + Using Ratios)

We know there is only one way to fit an equilateral triangle into a square: one of its vertices is a corner of the square and the other two vertices fall on opposite sides (try to imagine it in your head). It must be symmetrical along a diagonal of the square.

Thus $\triangle CNB \sim\triangle CDM$ where both are $15^{\circ}-75^{\circ}-90^{\circ}$ triangles since $\angle MCN=60^{\circ}$

Using the $15^{\circ}-75^{\circ}-90^{\circ}$ ratios ($2-\sqrt3:1:\sqrt6-\sqrt2$), we have $CN = \sqrt6 - \sqrt2$.

Thus $[\triangle CMN] = \frac{\sqrt3}{4}(\sqrt6 - \sqrt2)^2 = \frac{\sqrt3}{4}(8-4\sqrt3) = 2\sqrt3-3 \Rightarrow \fbox{A}$.

$\sim$ Jonysun

See Also

1974 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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
All AHSME Problems and Solutions

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