Difference between revisions of "1986 AIME Problems/Problem 9"

Revision as of 23:29, 22 October 2017

Problem

In $\triangle ABC$ , $AB= 425$ , $BC=450$ , and $AC=510$ . An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle. If these three segments are of an equal length $d$ , find $d$ .

Solution

Solution 1

$[asy] size(200); pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10); pair C=(0,0),A=(510,0),B=IP(circle(C,450),circle(A,425)); /* construct remaining points */ pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C); pair D=IP(Ea--(Ea+A-C),A--B),F=IP(Da--(Da+C-B),A--C),Fa=IP(E--(E+A-B),A--C); /* Construct P */ pair P = IP(D--Ea,E--Fa); dot(MP("P",P,NE)); pair X = IP(L(A,P,4), B--C); dot(MP("X",X,NW)); pair Y = IP(L(B,P,4), C--A); dot(MP("Y",Y,NE)); pair Z = IP(L(C,P,4), A--B); dot(MP("Z",Z,N)); D(A--X); D(B--Y); D(C--Z); D(MP("A",A,s)--MP("B",B,N,s)--MP("C",C,s)--cycle); MP("450",(B+C)/2,NW);MP("425",(A+B)/2,NE);MP("510",(A+C)/2); [/asy]$

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 2

$[asy] size(200); pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10); pair C=(0,0),A=(510,0),B=IP(circle(C,450),circle(A,425)); /* construct remaining points */ pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C); pair D=IP(Ea--(Ea+A-C),A--B),F=IP(Da--(Da+C-B),A--C),Fa=IP(E--(E+A-B),A--C); D(MP("A",A,s)--MP("B",B,N,s)--MP("C",C,s)--cycle); dot(MP("D",D,NE,s));dot(MP("E",E,NW,s));dot(MP("F",F,s));dot(MP("D'",Da,NE,s));dot(MP("E'",Ea,NW,s));dot(MP("F'",Fa,s)); D(D--Ea);D(Da--F);D(Fa--E); MP("450",(B+C)/2,NW);MP("425",(A+B)/2,NE);MP("510",(A+C)/2); /*P copied from above solution*/ pair P = IP(D--Ea,E--Fa); dot(MP("P",P,N)); [/asy]$

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.

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 3

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 $\frac {f + d + b}{A} + 1 = x^2*\left(\frac {1}{BC^2} + \frac {1}{AC^2} + \frac {1}{AB^2}\right)$

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 $1 + \sum_{cyc}\left(1 - \frac {x}{AB}\right) - \frac {x^2}{AB^2} = 0$ $1 + \sum_{cyc}1 - 2*\frac {x}{AB} = 0$ $\frac {2}{\frac {1}{AB} + \frac {1}{BC} + \frac {1}{CA}} = x$ answer follows after some hideous computation.

Solution 4

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 5

Refer to the diagram from Solution 2. 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 $18d-18x=450*17-17d$ . Using the previous equation to substitute for $x$ , we have: $18d-3*2550+15d=450*17-17d$ 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 $x$ .)

@@ Line 64: / Line 64: @@
 Let the length of the segment be <math>x</math> and the area of the triangle be <math>A</math>, using the theorem, we get:
-<math>\frac {c + e + d}{A} = \left(\frac {x}{BC}\right)^2</math>, <math>\frac {b + c + d}{A}= \left(\frac {x}{AC}\right)^2</math>, <math>\frac {a + b + f}{A} = \left(\frac {x}{AB}\right)^2</math>
+<math>\frac {d + e + f}{A} = \left(\frac {x}{BC}\right)^2</math>, <math>\frac {b + c + d}{A}= \left(\frac {x}{AC}\right)^2</math>, <math>\frac {a + b + f}{A} = \left(\frac {x}{AB}\right)^2</math>
 adding all these together and using <math>a + b + c + d + e + f = A</math> we get
 <math>\frac {f + d + b}{A} + 1 = x^2*\left(\frac {1}{BC^2} + \frac {1}{AC^2} + \frac {1}{AB^2}\right)</math>

1986 AIME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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