AIME 1986 · 第 9 题

Solution

Solution 1

Let the points at which the segments hit the triangle be called $D, D', E, E', F, F'$ as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are similar ($\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF$). The remaining three sections are parallelograms.

By similar triangles, $BE'=\frac{d}{510}\cdot450=\frac{15}{17}d$ and $EC=\frac{d}{425}\cdot450=\frac{18}{17}d$. Since $FD'=BC-EE'$, we have $900-\frac{33}{17}d=d$, so $d=\boxed{306}$.

Solution 2

Construct cevians $AX$, $BY$ and $CZ$ through $P$. Place masses of $x,y,z$ on $A$, $B$ and $C$ respectively; then $P$ has mass $x+y+z$.

Notice that $Z$ has mass $x+y$. On the other hand, by similar triangles, $\frac{CP}{CZ} = \frac{d}{AB}$. Hence by mass points we find that

$$\frac{x+y}{x+y+z} = \frac{d}{AB}$$ Similarly, we obtain

$$\frac{y+z}{x+y+z} = \frac{d}{BC} \qquad \text{and} \qquad \frac{z+x}{x+y+z} = \frac{d}{CA}$$ Summing these three equations yields

$$\frac{d}{AB} + \frac{d}{BC} + \frac{d}{CA} = \frac{x+y}{x+y+z} + \frac{y+z}{x+y+z} + \frac{z+x}{x+y+z} = \frac{2x+2y+2z}{x+y+z} = 2$$ Hence,

$d = \frac{2}{\frac{1}{AB} + \frac{1}{BC} + \frac{1}{CA}} = \frac{2}{\frac{1}{510} + \frac{1}{450} + \frac{1}{425}} = \frac{10}{\frac{1}{85}+\frac{1}{90}+\frac{1}{102}}$$= \frac{10}{\frac{1}{5}\left(\frac{1}{17}+\frac{1}{18}\right)+\frac{1}{102}}=\frac{10}{\frac{1}{5}\cdot\frac{35}{306}+\frac{3}{306}}=\frac{10}{\frac{10}{306}} = \boxed{306}$

Solution 3

Let the points at which the segments hit the triangle be called $D, D', E, E', F, F'$ as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are similar ($\triangle ABC \sim \triangle DD'P \sim \triangle PEE' \sim \triangle F'PF$). The remaining three sections are parallelograms.

Since $PDAF'$ is a parallelogram, we find $PF' = AD$, and similarly $PE = BD'$. So $d = PF' + PE = AD + BD' = 425 - DD'$. Thus $DD' = 425 - d$. By the same logic, $EE' = 450 - d$.

Since $\triangle DPD' \sim \triangle ABC$, we have the proportion:

$\frac{425-d}{425} = \frac{PD}{510} \Longrightarrow PD = 510 - \frac{510}{425}d = 510 - \frac{6}{5}d$

Doing the same with $\triangle PEE'$, we find that $PE' =510 - \frac{17}{15}d$. Now, $d = PD + PE' = 510 - \frac{6}{5}d + 510 - \frac{17}{15}d \Longrightarrow d\left(\frac{50}{15}\right) = 1020 \Longrightarrow d = \boxed{306}$.

Solution 4

Define the points the same as above.

Let $[CE'PF] = a$, $[E'EP] = b$, $[BEPD'] = c$, $[D'PD] = d$, $[DAF'P] = e$ and $[F'D'P] = f$

The key theorem we apply here is that the ratio of the areas of 2 similar triangles is the ratio of a pair of corresponding sides squared.

Let the length of the segment be $x$ and the area of the triangle be $A$, using the theorem, we get:

$\frac {d + e + f}{A} = \left(\frac {x}{BC}\right)^2$, $\frac {b + c + d}{A}= \left(\frac {x}{AC}\right)^2$, $\frac {a + b + f}{A} = \left(\frac {x}{AB}\right)^2$. Adding all these together and using $a + b + c + d + e + f = A$ we get $\frac {f + d + b}{A} + 1 = x^2 \cdot \left(\frac {1}{BC^2} + \frac {1}{AC^2} + \frac {1}{AB^2}\right)$

Using corresponding angles from parallel lines, it is easy to show that $\triangle ABC \sim \triangle F'PF$; since $ADPF'$ and $CFPE'$ are parallelograms, it is easy to show that $FF' = AC - x$

Now we have the side length ratio, so we have the area ratio $\frac {f}{A} = \left(\frac {AC - x}{AC}\right)^2 = \left(1 - \frac {x}{AC}\right)^2$. By symmetry, we have $\frac {d}{A} = \left(1 - \frac {x}{AB}\right)^2$ and $\frac {b}{A} = \left(1 - \frac {x}{BC}\right)^2$

Substituting these into our initial equation, we have $1 + \sum_{cyc}\left(1 - \frac {x}{AB}\right) - \frac {x^2}{AB^2} = 0$ $\implies 1 + \sum_{cyc}1 - 2 \cdot \frac {x}{AB} = 0$ $\implies \frac {2}{\frac {1}{AB} + \frac {1}{BC} + \frac {1}{CA}} = x$ and the answer follows after some hideous computation.

Solution 5

Refer to the diagram in solution 2; let $a^2=[E'EP]$, $b^2=[D'DP]$, and $c^2=[F'FP]$. Now, note that $[E'BD]$, $[D'DP]$, and $[E'EP]$ are similar, so through some similarities we find that $\frac{E'P}{PD}=\frac{a}{b}\implies\frac{E'D}{PD}=\frac{a+b}{b}\implies[E'BD]=b^2\left(\frac{a+b}{b}\right)^2=(a+b)^2$. Similarly, we find that $[D'AF]=(b+c)^2$ and $[F'CE]=(c+a)^2$, so $[ABC]=(a+b+c)^2$. Now, again from similarity, it follows that $\frac{d}{510}=\frac{a+b}{a+b+c}$, $\frac{d}{450}=\frac{b+c}{a+b+c}$, and $\frac{d}{425}=\frac{c+a}{a+b+c}$, so adding these together, simplifying, and solving gives $d=\frac{2}{\frac{1}{425}+\frac{1}{450}+\frac{1}{510}}=\frac{10}{\frac{1}{85}+\frac{1}{90}+\frac{1}{102}}=\frac{10}{\frac{1}{5}\left(\frac{1}{17}+\frac{1}{18}\right)+\frac{1}{102}}=\frac{10}{\frac{1}{5}\cdot\frac{35}{306}+\frac{3}{306}}$ $=\frac{10}{\frac{10}{306}}=\boxed{306}$.

Solution 6

Refer to the diagram above. Notice that because $CE'PF$, $AF'PD$, and $BD'PE$ are parallelograms, $\overline{DD'} = 425-d$, $\overline{EE'} = 450-d$, and $\overline{FF'} = 510-d$.

Let $F'P = x$. Then, because $\triangle ABC \sim \triangle F'PF$, $\frac{AB}{AC}=\frac{F'P}{F'F}$, so $\frac{425}{510}=\frac{x}{510-d}$. Simplifying the LHS and cross-multiplying, we have $6x=2550-5d$. From the same triangles, we can find that $FP=\frac{18}{17}x$.

$\triangle PEE'$ is also similar to $\triangle F'PF$. Since $EF'=d$, $EP=d-x$. We now have $\frac{PE}{EE'}=\frac{F'P}{FP}$, and $\frac{d-x}{450-d}=\frac{17}{18}$. Cross multiplying, we have $18d-18x=450 \cdot 17-17d$. Using the previous equation to substitute for $x$, we have:

$$18d-3\cdot2550+15d=450\cdot17-17d$$ This is a linear equation in one variable, and we can solve to get $d=\boxed{306}$

• I did not show the multiplication in the last equation because most of it cancels out when solving.

(Note: I chose $F'P$ to be $x$ only because that is what I had written when originally solving. The solution would work with other choices for $x$.)

Video Solution by Pi Academy

https://youtu.be/fHFD0TEfBnA?si=DRKVU_As7Rv0ou5D

~ Pi Academy