AIME 1998 · 第 12 题

Solution 1

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