AIME 1996 · 第 13 题

Solution

Let $E$ be the midpoint of $\overline{BC}$. Since $BE = EC$, then $\triangle ABE$ and $\triangle AEC$ share the same height and have equal bases, and thus have the same area. Similarly, $\triangle BDE$ and $BAE$ share the same height, and have bases in the ratio $DE : AE$, so $\frac{[BDE]}{[BAE]} = \frac{DE}{AE}$ (see area ratios). Now,

$\dfrac{[ADB]}{[ABC]} = \frac{[ABE] + [BDE]}{2[ABE]} = \frac{1}{2} + \frac{DE}{2AE}.$

By Stewart's Theorem, $AE = \frac{\sqrt{2(AB^2 + AC^2) - BC^2}}2 = \frac{\sqrt {57}}{2}$, and by the Pythagorean Theorem on $\triangle ABD, \triangle EBD$,

$$\begin{aligned} BD^2 + \left(DE + \frac {\sqrt{57}}2\right)^2 &= 30 \\ BD^2 + DE^2 &= \frac{15}{4} \\ \end{aligned}$$ Subtracting the two equations yields $DE\sqrt{57} + \frac{57}{4} = \frac{105}{4} \Longrightarrow DE = \frac{12}{\sqrt{57}}$. Then $\frac mn = \frac{1}{2} + \frac{DE}{2AE} = \frac{1}{2} + \frac{\frac{12}{\sqrt{57}}}{2 \cdot \frac{\sqrt{57}}{2}} = \frac{27}{38}$, and $m+n = \boxed{065}$.

Solution 2

Because the problem asks for a ratio, we can divide each side length by $\sqrt{3}$ to make things simpler. We now have a triangle with sides $\sqrt{10}$, $\sqrt{5}$, and $\sqrt{2}$.

We use the same graph as above.

Draw perpendicular from $C$ to $AE$. Denote this point as $F$. We know that $DE = EF = x$ and $BD = CF = z$ and also let $AE = y$.

Using Pythagorean theorem, we get three equations,

$(y+x)^2 + z^2 = 10$ $(y-x)^2 + z^2 = 2$ $x^2 + z^2 = \frac{5}{4}$

Adding the first and second, we obtain $x^2 + y^2 + z^2 = 6$, and then subtracting the third from this we find that $y = \frac{\sqrt{19}}{2}$. (Note, we could have used Stewart's Theorem to achieve this result).

Subtracting the first and second, we see that $xy = 2$, and then we find that $x = \frac{4}{\sqrt{19}}$

Using base ratios, we then quickly find that the desired ratio is $\frac{27}{38}$ so our answer is $\boxed{065}$

Solution 3 (Trig Bash)

First, extend side $AC$ to a point $F$ such that $AF \perp BF$.

We begin by using the Law of Cosines to find $\cos \angle ACB$:

$$(\sqrt{6})^2 + (\sqrt{15})^2 - 2(\sqrt{6})(\sqrt{15})\cos \angle ACB = (\sqrt{30})^2.$$ Simplifying,

$$6 + 15 - 2\sqrt{90}\cos \angle ACB = 30,$$ which gives

$$-6\sqrt{10}\cos \angle ACB = 9 \quad \Rightarrow \quad \cos \angle ACB = -\frac{3}{2\sqrt{10}}.$$ Using the Pythagorean identity,

$$\sin \angle ACB = \sqrt{1 - \cos^2 \angle ACB} = \frac{\sqrt{310}}{20}.$$ Let $[ABC]$ denote the area of $\triangle ABC$. Using the area formula $\tfrac{1}{2}ab\sin C$,

$$[ABC] = \frac{1}{2}(\sqrt{6})(\sqrt{15})\left(\frac{\sqrt{310}}{20}\right) = \frac{3\sqrt{31}}{4}.$$ Next, we use the Law of Cosines to find $AE$:

$$AE^2 = \left(\frac{\sqrt{15}}{2}\right)^2 + (\sqrt{6})^2 - 2\left(\frac{\sqrt{15}}{2}\right)(\sqrt{6})\left(-\frac{3}{2\sqrt{10}}\right).$$ Simplifying,

$$AE^2 = \frac{39}{4} + \frac{9}{2} = \frac{57}{4},$$ so

$$AE = \frac{\sqrt{57}}{2}.$$ Since $E$ is the midpoint of $BC$, we have $[ACE] = [ABE] = \frac{3\sqrt{31}}{8}$.

Solving for $\sin \angle BEA$,

$$\frac{1}{2}\left(\frac{\sqrt{15}}{2}\right)\left(\frac{\sqrt{57}}{2}\right)\sin \angle BEA = \frac{3\sqrt{31}}{8},$$ which simplifies to

$$\sqrt{95}\sin \angle BEA = \sqrt{31},$$ so

$$\sin \angle BEA = \sqrt{\frac{31}{95}}.$$ Using a Pythagorean identity,

$$\cos \angle BEA = \frac{8}{\sqrt{95}}.$$ Now, solving for $DE$,

$$\frac{DE}{\frac{\sqrt{15}}{2}} = \frac{8}{\sqrt{95}} \quad \Rightarrow \quad DE = \frac{4\sqrt{15}}{\sqrt{95}} = 4\sqrt{\frac{3}{19}}.$$ We now find the area of $\triangle BDE$:

$$[BDE] = \frac{1}{2}\left(4\sqrt{\frac{3}{19}}\right)\left(\frac{\sqrt{15}}{2}\right)\left(\sqrt{\frac{31}{95}}\right) = \frac{6\sqrt{31}}{38}.$$ Adding areas,

$$[ADB] = [ABE] + [BDE] = \frac{3\sqrt{31}}{8} + \frac{6\sqrt{31}}{38} = \frac{81\sqrt{31}}{152}.$$ Taking the ratio,

$$\frac{[ADB]}{[ABC]} = \frac{\frac{81\sqrt{31}}{152}}{\frac{3\sqrt{31}}{4}} = \frac{27}{38}.$$ Since the problem asks for $m+n$, we have $27 + 38 = \boxed{065}.$

~Voidling