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

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m (Added simplification from 4th solution to 1st solution)
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Notice that <math>Z</math> has mass <math>x+y</math>.  On the other hand, by similar triangles, <math>\frac{CP}{CZ} = \frac{d}{AB}</math>.  Hence by mass points we find that <cmath> \frac{x+y}{x+y+z} = \frac{d}{AB} </cmath> Similarly, we obtain <cmath> \frac{y+z}{x+y+z} = \frac{d}{BC} \qquad \text{and} \qquad \frac{z+x}{x+y+z} = \frac{d}{CA} </cmath> Summing these three equations yields <cmath> \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 </cmath>
 
Notice that <math>Z</math> has mass <math>x+y</math>.  On the other hand, by similar triangles, <math>\frac{CP}{CZ} = \frac{d}{AB}</math>.  Hence by mass points we find that <cmath> \frac{x+y}{x+y+z} = \frac{d}{AB} </cmath> Similarly, we obtain <cmath> \frac{y+z}{x+y+z} = \frac{d}{BC} \qquad \text{and} \qquad \frac{z+x}{x+y+z} = \frac{d}{CA} </cmath> Summing these three equations yields <cmath> \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 </cmath>
  
−
Hence, <cmath> d = \frac{2}{\frac{1}{AB} + \frac{1}{BC} + \frac{1}{CA}} = \frac{2}{\frac{1}{510} + \frac{1}{450} + \frac{1}{425}} = \boxed{306} </cmath>
+
Hence, <center><math> 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}}</math><math>=\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}</math></center>
 +
 
  
 
=== Solution 2 ===
 
=== Solution 2 ===

Revision as of 21:43, 13 March 2015

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 (mass points)

[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:

$\frac {c + e + d}{A} = \left(\frac {x}{BC}\right)^2$, $\frac {b + c + d}{A}= \left(\frac {x}{AC}\right)^2$, $\frac {a + b + f}{A} = \left(\frac {x}{AB}\right)^2$ 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 $d=\frac{2}{\frac{1}{425}+\frac{1}{450}+\frac{1}{510}}=\frac{10}{\frac{1}{85}+\frac{1}{90}+\frac{1}{102}}=\frac{10}{\frac{1}{5}\left(\frac{1}{17}+\frac{1}{18}\right)+\frac{1}{102}}=\frac{10}{\frac{1}{5}\cdot\frac{35}{306}+\frac{3}{306}}$ $=\frac{10}{\frac{10}{306}}=\boxed{306}$.

See also

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