AIME 2006 II · 第 15 题

Solution 1 (Geometric Interpretation)

Let $\triangle XYZ$ be a triangle with sides of length $x, y$ and $z$, and suppose this triangle is acute (so all altitudes are in the interior of the triangle).

Let the altitude to the side of length $x$ be of length $h_x$, and similarly for $y$ and $z$. Then we have by two applications of the Pythagorean Theorem we that

$$x = \sqrt{y^2 - h_x^2} + \sqrt{z^2 - h_x^2}$$ As a function of $h_x$, the RHS of this equation is strictly decreasing, so it takes each value in its range exactly once. Thus we must have that $h_x^2 = \frac1{16}$ and so $h_x = \frac{1}4$ and similarly $h_y = \frac15$ and $h_z = \frac16$.

The area of the triangle must be the same no matter how we measure it; therefore $x\cdot h_x = y\cdot h_y = z \cdot h_z$ gives us $\frac x4 = \frac y5 = \frac z6 = 2A$ and $x = 8A, y = 10A$ and $z = 12A$.

The semiperimeter of the triangle is $s = \frac{8A + 10A + 12A}{2} = 15A$ so by Heron's formula we have

$$A = \sqrt{15A \cdot 7A \cdot 5A \cdot 3A} = 15A^2\sqrt{7}$$ Thus, $A = \frac{1}{15\sqrt{7}}$ and $x + y + z = 30A = \frac2{\sqrt{7}}$ and the answer is $2 + 7 = \boxed{009}$.

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The justification that there is an acute triangle with sides of length $x, y$ and $z$:

Note that $x, y$ and $z$ are each the sum of two positive square roots of real numbers, so $x, y, z \geq 0$. (Recall that, by AIME convention, all numbers (including square roots) are taken to be real unless otherwise indicated.)

Also, $\sqrt{y^2-\frac{1}{16}} < \sqrt{y^2} = y$, so we have $x < y + z$, $y < z + x$ and $z < x + y$. But these conditions are exactly those of the triangle inequality, so there does exist such a triangle.

Solution 2 (Heron Bash)

We can rewrite the equations as follows:

$$\begin{aligned} \frac{1}{4}x &= \sqrt{\left(y+\frac{1}{4}\right) \left(y-\frac{1}{4}\right)\left(\frac{1}{4}\right) \left(\frac{1}{4}\right)}+\sqrt{\left(z+\frac{1}{4}\right) \left(z-\frac{1}{4}\right) \left(\frac{1}{4}\right) \left(\frac{1}{4}\right)} \\ \frac{1}{5} y &= \sqrt{\left(z+\frac{1}{5}\right) \left(z-\frac{1}{5}\right) \left(\frac{1}{5}\right) \left(\frac{1}{5}\right)}+\sqrt{\left(x+\frac{1}{5}\right) \left(x-\frac{1}{5}\right) \left(\frac{1}{5}\right) \left(\frac{1}{5}\right)} \\ \frac{1}{6} z &= \sqrt{\left(x+\frac{1}{6}\right) \left(x-\frac{1}{6}\right) \left(\frac{1}{6}\right) \left(\frac{1}{6}\right)}+\sqrt{\left(y+\frac{1}{6}\right) \left(y-\frac{1}{6}\right) \left(\frac{1}{6}\right) \left(\frac{1}{6}\right)} \end{aligned}$$ Take the first equation. The first square root is the area of a triangle with side lengths $y,y,\frac{1}{2}$ by Heron’s Formula. Similarly, the second square root is the area of a triangle with side lengths $z,z,\frac{1}{2}$. If we connect these two triangles together at the $\frac{1}{2}$ side, we obtain a kite. The area of the kite is $\frac{1}{4}x$, and since the first diagonal is $\frac{1}{2}$, the second diagonal is $\frac{2\cdot\frac{1}{4}x}{\frac{1}{2}}=x$. If we draw this diagonal, we obtain two triangles with side lengths $x,y,z$. Let this triangle have area $A$. Then $\frac{1}{4}x=2A$; extend this for the other two equations to get $\frac{1}{5}y=2A$ and $\frac{1}{6}z=2A$. Therefore $x=8A$, $y=10A$, and $z=12A$. But we can find the area of the $x,y,z$ triangle with Heron's:

$$A=\sqrt{(15A)(7A)(5A)(3A)}=15A^2\sqrt{7}$$ Therefore $A=\frac{1}{15\sqrt{7}}$. Hence $x+y+z=30A=\frac{2}{\sqrt{7}}\Rightarrow\boxed{009}$.

~eevee9406

Solution 3 (Algebraic)

Note that none of $x,y,z$ can be zero.

Each of the equations is in the form

$$a=\sqrt{b^2-d^2}+\sqrt{c^2-d^2}$$ Isolate a radical and square the equation to get

$$b^2-d^2=a^2-2a\sqrt{c^2-d^2}+c^2-d^2$$ Now cancel, and again isolate the radical, and square the equation to get

$$a^4+b^4+c^4+2a^2c^2-2a^2b^2-2b^2c^2=4a^2c^2-4a^2d^2$$ Rearranging gives

$$a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2-4a^2d^2$$ Now note that everything is cyclic but the last term (i.e. $-4a^2d^2$), which implies

$$-4x^2\cdot\frac1{16}=-4y^2\cdot\frac1{25}=-4z^2\cdot\frac1{36}$$ Or

$$x: y: z=4: 5: 6 \implies x=\frac{4y}5 \textrm{ and } z=\frac{6y}5$$ Plug these values into the middle equation to get

$$\frac{256y^4+625y^4+1296y^4}{625}=\frac{800y^4}{625}+\frac{1800y^4}{625}+\frac{1152y^4}{625}-\frac{100y^2}{625}$$ Simplifying gives

$$1575y^4=100y^2 \textrm{ but } y \neq 0 \implies y^2=\frac4{63} \textrm{ or } y=\frac2{3\sqrt7}$$ Substituting the value of $y$ for $x$ and $z$ gives

$$x+y+z = \frac{4y+5y+6y}5 = 3y = 3 \cdot \frac{2}{3\sqrt7} = \frac{2}{\sqrt7}$$ And thus the answer is $\boxed{009}$

~phoenixfire

Video solution

https://www.youtube.com/watch?v=M6sC26dzb_I

Video Solution by OmegaLearn.org

https://youtu.be/9Vlt9g8TgRs