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Difference between revisions of "1998 AIME Problems/Problem 12"

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[[Image:1998_AIME-12.png]]
 
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WLOG, assume that <math>AB = BC = AC = 2</math>.
  
 
We let <math>x = EP = FQ</math>, <math>y = EQ</math>, <math>k = PQ</math>. Since <math>AE = \frac {1}{2}AB</math> and <math>AD = \frac {1}{2}AC</math>, <math>\triangle AED \sim \triangle ABC</math> and <math>ED \parallel BC</math>.
 
We let <math>x = EP = FQ</math>, <math>y = EQ</math>, <math>k = PQ</math>. Since <math>AE = \frac {1}{2}AB</math> and <math>AD = \frac {1}{2}AC</math>, <math>\triangle AED \sim \triangle ABC</math> and <math>ED \parallel BC</math>.
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We want the ratio of the squares of the sides, so <math>\frac {(2)^{2}}{k^{2}} = \frac {4}{7 - 3\sqrt {5}} = 7 + 3\sqrt {5}</math> so <math>a^{2} + b^{2} + c^{2} = 7^{2} + 3^{2} + 5^{2} = \boxed{083}</math>.
 
We want the ratio of the squares of the sides, so <math>\frac {(2)^{2}}{k^{2}} = \frac {4}{7 - 3\sqrt {5}} = 7 + 3\sqrt {5}</math> so <math>a^{2} + b^{2} + c^{2} = 7^{2} + 3^{2} + 5^{2} = \boxed{083}</math>.
  
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== Solution 2 in progress ==
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== Solution 2 ==
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WLOG, let <math>\Delta ABC</math> have side length <math>2.</math> Then, <math>DE = EF = FD = 1.</math> We also notice that <math>\angle CEP = \angle DEF = 60^{\circ},</math> meaning <math>\angle CEF = \angle CEP + \angle DEF = 120^{\circ}.</math>
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Let <math>EP = x.</math> Since <math>FQ = x</math> by congruent triangles <math>\Delta EPC</math> and <math>\Delta FQA,</math> <math>EQ = EF - FQ = 1-x.</math> We can now apply Law of Cosines to <math>\Delta CEP, \Delta PEQ,</math> and <math>\Delta CEQ.</math>
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By LoC on <math>\Delta CEP,</math> we get <cmath>CP^2 = 1^2 + x^2 - 2\cdot 1\cdot x\cdot \left(\frac{1}{2}\right) = x^2 - x + 1.</cmath>
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In a similar vein, using LoC on <math>\Delta PEQ</math> and <math>\Delta CEQ,</math> respectively, earns <cmath>PQ^2 = x^2 + (1-x)^2 - 2\cdot x\cdot (1-x)\cdot \left(\frac{1}{2}\right) = 3x^2 - 3x + 1</cmath> <cmath>CQ^2 = 1^2 + (1-x)^2 - 2\cdot 1\cdot (1-x)\cdot \left(-\frac{1}{2}\right) = x^2 - 3x + 3</cmath>
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We have <math>CP^2, PQ^2,</math> and <math>CQ^2.</math> Additionally, by segment addition, <math>CP + PQ = CQ.</math> Solving for <math>CP, PQ,</math> and <math>CQ</math> from the Law of Cosines expressions and plugging them into the segment addition gets the (admittedly-ugly) equation <cmath>\sqrt{x^2-3x+3} = \sqrt{x^2-x+1} + \sqrt{3x^2-3x+1}.</cmath>
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Since the equation is ugly, we look at what the problem is asking for us to solve. We want <math>\frac{[ABC]}{[PQR]}.</math> We see that <math>[ABC] = \sqrt{3}</math> and <math>[PQR] = [DEF] - [PDR] - [RFQ] - [QEP] = \frac{\sqrt{3}}{4} - \frac{3}{2}\left(\frac{\sqrt{3}}{2}x(1-x)\right),</math> since <math>[PDR] = [RFQ] = [QEP] = \frac{1}{2}x(1-x)\frac{\sqrt{3}}{2}</math> from the sine area formula. Simplifying <math>\frac{[ABC]}{[PQR]}</math> gets us wanting to find <math>\frac{4}{3x^2-3x+1}.</math>
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We see <math>3x^2-3x+1</math> in both the denominator of what we want and under a radicand in our algebraic expression, which leads us to think the calculations may not be that bad. Isolate <math>\sqrt{3x^2-3x+1}</math> and square to get <cmath>3x^2-3x+1 = 2x^2-4x+4-2\sqrt{(x^2-3x+3)(x^2-x+1)}</cmath>
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Isolate the radicand and square and expand to get <math>x^4+2x^3-5x^2-6x+9=4x^4-16x^3+28x^2-24x+12,</math> and moving terms to one side and dividing by <math>3,</math> we get <cmath>x^4-6x^3+11x^2-6x+1=0.</cmath>
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This can be factored into <math>(x^2-3x+1)^2 = 0 \rightarrow x^2-3x+1 = 0 \rightarrow x = \frac{3 \pm \sqrt{5}}{2}.</math> From the equation <math>x^2-3x+1=0,</math> we have <math>x^2=3x+1,</math> so plugging that value into the expression we want to find, we get <math>\frac{4}{3(3x+1)-3x+1} = \frac{4}{6x+2}.</math>
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Substituting <math>x = \frac{3-\sqrt{5}}{2}</math> into <math>\frac{4}{6x+2}</math> gets an expression of <math>7+3\sqrt{5},</math> so <math>a^2+b^2+c^2 = \boxed{083}</math>.
  
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[[Image:1998_AIME-12.png]]
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-PureSwag
  
 
== See also ==
 
== See also ==

Latest revision as of 13:33, 18 June 2024

Problem

Let $ABC$ be equilateral, and $D, E,$ and $F$ be the midpoints of $\overline{BC}, \overline{CA},$ and $\overline{AB},$ respectively. There exist points $P, Q,$ and $R$ on $\overline{DE}, \overline{EF},$ and $\overline{FD},$ respectively, with the property that $P$ is on $\overline{CQ}, Q$ is on $\overline{AR},$ and $R$ is on $\overline{BP}.$ The ratio of the area of triangle $ABC$ to the area of triangle $PQR$ is $a + b\sqrt {c},$ where $a, b$ and $c$ are integers, and $c$ is not divisible by the square of any prime. What is $a^{2} + b^{2} + c^{2}$?

Solution 1

1998 AIME-12.png

WLOG, assume that $AB = BC = AC = 2$.

We let $x = EP = FQ$, $y = EQ$, $k = PQ$. Since $AE = \frac {1}{2}AB$ and $AD = \frac {1}{2}AC$, $\triangle AED \sim \triangle ABC$ and $ED \parallel BC$.

By alternate interior angles, we have $\angle PEQ = \angle BFQ$ and $\angle EPQ = \angle FBQ$. By vertical angles, $\angle EQP = \angle FQB$.

Thus $\triangle EQP \sim \triangle FQB$, so $\frac {EP}{EQ} = \frac {FB}{FQ}\Longrightarrow\frac {x}{y} = \frac {1}{x}\Longrightarrow x^{2} = y$.

Since $\triangle EDF$ is equilateral, $EQ + FQ = EF = BF = 1\Longrightarrow x + y = 1$. Solving for $x$ and $y$ using $x^{2} = y$ and $x + y = 1$ gives $x = \frac {\sqrt {5} - 1}{2}$ and $y = \frac {3 - \sqrt {5}}{2}$.

Using the Law of Cosines, we get

$k^{2} = x^{2} + y^{2} - 2xy\cos{\frac {\pi}{3}}$
$= \left(\frac {\sqrt {5} - 1}{2}\right)^{2} + \left(\frac {3 - \sqrt {5}}{2}\right)^{2} - 2\left(\frac {\sqrt {5} - 1}{2}\right)\left(\frac {3 - \sqrt {5}}{2}\right)\cos{\frac {\pi}{3}}$
$= 7 - 3\sqrt {5}$

We want the ratio of the squares of the sides, so $\frac {(2)^{2}}{k^{2}} = \frac {4}{7 - 3\sqrt {5}} = 7 + 3\sqrt {5}$ so $a^{2} + b^{2} + c^{2} = 7^{2} + 3^{2} + 5^{2} = \boxed{083}$.

Solution 2

WLOG, let $\Delta ABC$ have side length $2.$ Then, $DE = EF = FD = 1.$ We also notice that $\angle CEP = \angle DEF = 60^{\circ},$ meaning $\angle CEF = \angle CEP + \angle DEF = 120^{\circ}.$

Let $EP = x.$ Since $FQ = x$ by congruent triangles $\Delta EPC$ and $\Delta FQA,$ $EQ = EF - FQ = 1-x.$ We can now apply Law of Cosines to $\Delta CEP, \Delta PEQ,$ and $\Delta CEQ.$

By LoC on $\Delta CEP,$ we get \[CP^2 = 1^2 + x^2 - 2\cdot 1\cdot x\cdot \left(\frac{1}{2}\right) = x^2 - x + 1.\]

In a similar vein, using LoC on $\Delta PEQ$ and $\Delta CEQ,$ respectively, earns \[PQ^2 = x^2 + (1-x)^2 - 2\cdot x\cdot (1-x)\cdot \left(\frac{1}{2}\right) = 3x^2 - 3x + 1\] \[CQ^2 = 1^2 + (1-x)^2 - 2\cdot 1\cdot (1-x)\cdot \left(-\frac{1}{2}\right) = x^2 - 3x + 3\]

We have $CP^2, PQ^2,$ and $CQ^2.$ Additionally, by segment addition, $CP + PQ = CQ.$ Solving for $CP, PQ,$ and $CQ$ from the Law of Cosines expressions and plugging them into the segment addition gets the (admittedly-ugly) equation \[\sqrt{x^2-3x+3} = \sqrt{x^2-x+1} + \sqrt{3x^2-3x+1}.\]

Since the equation is ugly, we look at what the problem is asking for us to solve. We want $\frac{[ABC]}{[PQR]}.$ We see that $[ABC] = \sqrt{3}$ and $[PQR] = [DEF] - [PDR] - [RFQ] - [QEP] = \frac{\sqrt{3}}{4} - \frac{3}{2}\left(\frac{\sqrt{3}}{2}x(1-x)\right),$ since $[PDR] = [RFQ] = [QEP] = \frac{1}{2}x(1-x)\frac{\sqrt{3}}{2}$ from the sine area formula. Simplifying $\frac{[ABC]}{[PQR]}$ gets us wanting to find $\frac{4}{3x^2-3x+1}.$

We see $3x^2-3x+1$ in both the denominator of what we want and under a radicand in our algebraic expression, which leads us to think the calculations may not be that bad. Isolate $\sqrt{3x^2-3x+1}$ and square to get \[3x^2-3x+1 = 2x^2-4x+4-2\sqrt{(x^2-3x+3)(x^2-x+1)}\]

Isolate the radicand and square and expand to get $x^4+2x^3-5x^2-6x+9=4x^4-16x^3+28x^2-24x+12,$ and moving terms to one side and dividing by $3,$ we get \[x^4-6x^3+11x^2-6x+1=0.\]

This can be factored into $(x^2-3x+1)^2 = 0 \rightarrow x^2-3x+1 = 0 \rightarrow x = \frac{3 \pm \sqrt{5}}{2}.$ From the equation $x^2-3x+1=0,$ we have $x^2=3x+1,$ so plugging that value into the expression we want to find, we get $\frac{4}{3(3x+1)-3x+1} = \frac{4}{6x+2}.$

Substituting $x = \frac{3-\sqrt{5}}{2}$ into $\frac{4}{6x+2}$ gets an expression of $7+3\sqrt{5},$ so $a^2+b^2+c^2 = \boxed{083}$.

-PureSwag

See also

1998 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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All AIME Problems and Solutions

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