AIME 2007 I · 第 13 题

Solution

Solution 1

Note first that the intersection is a pentagon.

Use 3D analytical geometry, setting the origin as the center of the square base and the pyramid’s points oriented as shown above. $A(-2,2,0),\ B(2,2,0),\ C(2,-2,0),\ D(-2,-2,0),\ E(0,0,2\sqrt{2})$. Using the coordinates of the three points of intersection $(-1,1,\sqrt{2}),\ (2,0,0),\ (0,-2,0)$, it is possible to determine the equation of the plane. The equation of a plane resembles $ax + by + cz = d$, and using the points we find that $2a = d \Longrightarrow d = \frac{a}{2}$, $-2b = d \Longrightarrow d = \frac{-b}{2}$, and $-a + b + \sqrt{2}c = d \Longrightarrow -\frac{d}{2} - \frac{d}{2} + \sqrt{2}c = d \Longrightarrow c = d\sqrt{2}$. It is then $x - y + 2\sqrt{2}z = 2$.

Write the equation of the lines and substitute to find that the other two points of intersection on $\overline{BE}$, $\overline{DE}$ are $\left(\frac{\pm 3}{2},\frac{\pm 3}{2},\frac{\sqrt{2}}{2}\right)$. To find the area of the pentagon, break it up into pieces (an isosceles triangle on the top, an isosceles trapezoid on the bottom). Using the distance formula ($\sqrt{a^2 + b^2 + c^2}$), it is possible to find that the area of the triangle is $\frac{1}{2}bh \Longrightarrow \frac{1}{2} 3\sqrt{2} \cdot \sqrt{\frac 52} = \frac{3\sqrt{5}}{2}$. The trapezoid has area $\frac{1}{2}h(b_1 + b_2) \Longrightarrow \frac 12\sqrt{\frac 52}\left(2\sqrt{2} + 3\sqrt{2}\right) = \frac{5\sqrt{5}}{2}$. In total, the area is $4\sqrt{5} = \sqrt{80}$, and the solution is $\boxed{080}$.

Solution 2

Use the same coordinate system as above, and let the plane determined by $\triangle PQR$ intersect $\overline{BE}$ at $X$ and $\overline{DE}$ at $Y$. Then the line $\overline{XY}$ is the intersection of the planes determined by $\triangle PQR$ and $\triangle BDE$.

Note that the plane determined by $\triangle BDE$ has the equation $x=y$, and $\overline{PQ}$ can be described by $x=2(1-t)-t,\ y=t,\ z=t\sqrt{2}$. It intersects the plane when $2(1-t)-t=t$, or $t=\frac{1}{2}$. This intersection point has $z=\frac{\sqrt{2}}{2}$. Similarly, the intersection between $\overline{PR}$ and $\triangle BDE$ has $z=\frac{\sqrt{2}}{2}$. So $\overline{XY}$ lies on the plane $z=\frac{\sqrt{2}}{2}$, from which we obtain $X=\left( \frac{3}{2},\frac{3}{2},\frac{\sqrt{2}}{2}\right)$ and $Y=\left( -\frac{3}{2},-\frac{3}{2},\frac{\sqrt{2}}{2}\right)$. The area of the pentagon $EXQRY$ can be computed in the same way as above.

Solution 3

Extend $\overline{RQ}$ and $\overline{AB}$. The point of intersection is $H$. Connect $\overline{PH}$. $\overline{EB}$ intersects $\overline{PH}$ at $X$. Do the same for $\overline{QR}$ and $\overline{AD}$, and let the intersections be $I$ and $Y$

Because $Q$ is the midpoint of $\overline{BC}$, and $\overline{AB}\parallel\overline{DC}$, so $\triangle{RQC}\cong\triangle{HQB}$. $\overline{BH}=2$.

Because $\overline{BH}=2$, we can use mass point geometry to get that $\overline{PX}=\overline{XH}$. $|\triangle{XHQ}|=\frac{\overline{XH}}{\overline{PH}}\cdot\frac{\overline{QH}}{\overline{HI}}\cdot|\triangle{PHI}|=\frac{1}{6}\cdot|\triangle{PHI}|$

Using the same principle, we can get that $|\triangle{IYR}|=\frac{1}{6}|\triangle{PHI}|$

Therefore, the area of $PYRQX$ is $\frac{2}{3}\cdot|\triangle{PHI}|$

$\overline{RQ}=2\sqrt{2}$, so $\overline{IH}=6\sqrt{2}$. Using the law of cosines, $\overline{PH}=\sqrt{28}$. The area of $\triangle{PHI}=\frac{1}{2}\cdot\sqrt{28-18}\cdot6\sqrt{2}=6\sqrt{5}$

Using this, we can get the area of $PYRQX = \sqrt{80}$ so the answer is $\fbox{080}$.

Video Solution

2007 AIME I #13

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