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

m (Added point P in second solution)
m (fixed diagram)
 
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__TOC__
 
__TOC__
 
== Solution ==
 
== Solution ==
−
===Solution 1 (mass points) ===
+
=== Solution 1 ===
 +
<center><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></center> <!-- Asymptote replacement for Image:1986_AIME-9.png by azjps -->
 +
 
 +
Let the points at which the segments hit the triangle be called <math>D, D', E, E', F, F'</math> as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are [[similar triangles|similar]] (<math>\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF</math>). The remaining three sections are [[parallelogram]]s.
 +
 
 +
By similar triangles, <math>BE'=\frac{d}{510}\cdot450=\frac{15}{17}d</math> and <math>EC=\frac{d}{425}\cdot450=\frac{18}{17}d</math>. Since <math>FD'=BC-EE'</math>, we have <math>900-\frac{33}{17}d=d</math>, so <math>d=\boxed{306}</math>.
 +
 
 +
===Solution 2 ===
 
<asy>
 
<asy>
 
size(200);
 
size(200);
Line 27: Line 47:
 
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 3 ===
 
<center><asy>
 
<center><asy>
 
size(200);
 
size(200);
−
pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10);
+
pathpen = black;
−
pair C=(0,0),A=(510,0),B=IP(circle(C,450),circle(A,425));
+
pointpen = black + linewidth(0.6);
−
/* construct remaining points */
+
pen s = fontsize(10);
−
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);  
+
// Define points
−
D(MP("A",A,s)--MP("B",B,N,s)--MP("C",C,s)--cycle);
+
pair C = (0,0), A = (510,0);
−
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));
+
pair B = IP(circle(C,450),circle(A,425));
−
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);
+
// Construct remaining points
−
/*P copied from above solution*/
+
pair Da = IP(circle(A,289),A--B);
−
pair P = IP(D--Ea,E--Fa); dot(MP("P",P,N));  
+
pair E = IP(circle(C,324),B--C);
 +
pair Ea = IP(circle(B,270),B--C);
 +
pair D = IP(Ea--(Ea+A-C),A--B);
 +
pair F = IP(Da--(Da+C-B),A--C);
 +
pair Fa = IP(E--(E+A-B),A--C);
 +
 
 +
// Draw the main triangle
 +
draw(A--B--C--cycle);
 +
dot(MP("A",A,s));
 +
dot(MP("B",B,N,s));
 +
dot(MP("C",C,s));
 +
 
 +
// Mark and draw the other points
 +
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));
 +
 
 +
// Draw connecting lines
 +
draw(D--Ea);
 +
draw(Da--F);
 +
draw(Fa--E);
 +
 
 +
// Label distances
 +
label("450", (B+C)/2, NW);
 +
label("425", (A+B)/2, NE);
 +
label("510", (A+C)/2, S);
 +
 
 +
// Additional point P
 +
pair P = IP(D--Ea, E--Fa);
 +
dot(MP("P",P,N));
 
</asy></center> <!-- Asymptote replacement for Image:1986_AIME-9.png by azjps -->
 
</asy></center> <!-- Asymptote replacement for Image:1986_AIME-9.png by azjps -->
  
−
Let the points at which the segments hit the triangle be called <math>D, D', E, E', F, F'</math> as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are [[similar triangles|similar]] (<math>\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF</math>). The remaining three sections are [[parallelogram]]s.
+
Let the points at which the segments hit the triangle be called <math>D, D', E, E', F, F'</math> as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are [[similar triangles|similar]] (<math>\triangle ABC \sim \triangle DD'P \sim \triangle PEE' \sim \triangle F'PF</math>). The remaining three sections are [[parallelogram]]s.
  
 
Since <math>PDAF'</math> is a parallelogram, we find <math>PF' = AD</math>, and similarly <math>PE = BD'</math>. So <math>d = PF' + PE = AD + BD' = 425 - DD'</math>. Thus <math>DD' = 425 - d</math>. By the same logic, <math>EE' = 450 - d</math>.  
 
Since <math>PDAF'</math> is a parallelogram, we find <math>PF' = AD</math>, and similarly <math>PE = BD'</math>. So <math>d = PF' + PE = AD + BD' = 425 - DD'</math>. Thus <math>DD' = 425 - d</math>. By the same logic, <math>EE' = 450 - d</math>.  
Line 55: Line 107:
 
Doing the same with <math>\triangle PEE'</math>, we find that <math>PE' =510 - \frac{17}{15}d</math>. Now, <math>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}</math>.
 
Doing the same with <math>\triangle PEE'</math>, we find that <math>PE' =510 - \frac{17}{15}d</math>. Now, <math>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}</math>.
  
−
=== Solution 3 ===
+
=== Solution 4 ===
 
Define the points the same as above.
 
Define the points the same as above.
  
Line 64: Line 116:
 
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:
 
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
+
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>
+
<math>\frac {f + d + b}{A} + 1 = x^2 \cdot \left(\frac {1}{BC^2} + \frac {1}{AC^2} + \frac {1}{AB^2}\right)</math>
  
−
Using [[corresponding angles]] from parallel lines, it is easy to show that <math>\triangle ABC \sim \triangle F'PF</math>, since <math>ADPF'</math> and <math>CFPE'</math> are parallelograms, it is easy to show that <math>FF' = AC - x</math>
+
Using [[corresponding angles]] from parallel lines, it is easy to show that <math>\triangle ABC \sim \triangle F'PF</math>; since <math>ADPF'</math> and <math>CFPE'</math> are parallelograms, it is easy to show that <math>FF' = AC - x</math>
  
 
Now we have the side length [[ratio]], so we have the area ratio
 
Now we have the side length [[ratio]], so we have the area ratio
−
<math>\frac {f}{A} = \left(\frac {AC - x}{AC}\right)^2 = \left(1 - \frac {x}{AC}\right)^2</math>, by symmetry, we have
+
<math>\frac {f}{A} = \left(\frac {AC - x}{AC}\right)^2 = \left(1 - \frac {x}{AC}\right)^2</math>. By symmetry, we have
 
<math>\frac {d}{A} = \left(1 - \frac {x}{AB}\right)^2</math> and <math>\frac {b}{A} = \left(1 - \frac {x}{BC}\right)^2</math>
 
<math>\frac {d}{A} = \left(1 - \frac {x}{AB}\right)^2</math> and <math>\frac {b}{A} = \left(1 - \frac {x}{BC}\right)^2</math>
  
 
Substituting these into our initial equation, we have
 
Substituting these into our initial equation, we have
 
<math>1 + \sum_{cyc}\left(1 - \frac {x}{AB}\right) - \frac {x^2}{AB^2} = 0</math>
 
<math>1 + \sum_{cyc}\left(1 - \frac {x}{AB}\right) - \frac {x^2}{AB^2} = 0</math>
−
<math>1 + \sum_{cyc}1 - 2*\frac {x}{AB} = 0</math>
+
<math>\implies 1 + \sum_{cyc}1 - 2 \cdot \frac {x}{AB} = 0</math>
−
<math>\frac {2}{\frac {1}{AB} + \frac {1}{BC} + \frac {1}{CA}} = x</math>
+
<math>\implies \frac {2}{\frac {1}{AB} + \frac {1}{BC} + \frac {1}{CA}} = x</math>
−
answer follows after some hideous computation.
+
and the answer follows after some hideous computation.
  
−
===Solution 4===
+
===Solution 5===
 
Refer to the diagram in solution 2; let <math>a^2=[E'EP]</math>, <math>b^2=[D'DP]</math>, and <math>c^2=[F'FP]</math>. Now, note that <math>[E'BD]</math>, <math>[D'DP]</math>, and <math>[E'EP]</math> are similar, so through some similarities we find that <math>\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</math>. Similarly, we find that <math>[D'AF]=(b+c)^2</math> and <math>[F'CE]=(c+a)^2</math>, so <math>[ABC]=(a+b+c)^2</math>. Now, again from similarity, it follows that <math>\frac{d}{510}=\frac{a+b}{a+b+c}</math>, <math>\frac{d}{450}=\frac{b+c}{a+b+c}</math>, and <math>\frac{d}{425}=\frac{c+a}{a+b+c}</math>, so adding these together, simplifying, and solving gives <math>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}}</math>
 
Refer to the diagram in solution 2; let <math>a^2=[E'EP]</math>, <math>b^2=[D'DP]</math>, and <math>c^2=[F'FP]</math>. Now, note that <math>[E'BD]</math>, <math>[D'DP]</math>, and <math>[E'EP]</math> are similar, so through some similarities we find that <math>\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</math>. Similarly, we find that <math>[D'AF]=(b+c)^2</math> and <math>[F'CE]=(c+a)^2</math>, so <math>[ABC]=(a+b+c)^2</math>. Now, again from similarity, it follows that <math>\frac{d}{510}=\frac{a+b}{a+b+c}</math>, <math>\frac{d}{450}=\frac{b+c}{a+b+c}</math>, and <math>\frac{d}{425}=\frac{c+a}{a+b+c}</math>, so adding these together, simplifying, and solving gives <math>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}}</math>
 
<math>=\frac{10}{\frac{10}{306}}=\boxed{306}</math>.
 
<math>=\frac{10}{\frac{10}{306}}=\boxed{306}</math>.
 +
 +
=== Solution 6 ===
 +
<center><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></center> <!-- Asymptote replacement for Image:1986_AIME-9.png by azjps -->
 +
Refer to the diagram above. Notice that because <math>CE'PF</math>, <math>AF'PD</math>, and <math>BD'PE</math> are parallelograms, <math>\overline{DD'} = 425-d</math>, <math>\overline{EE'} = 450-d</math>, and <math>\overline{FF'} = 510-d</math>.
 +
 +
Let <math>F'P = x</math>. Then, because <math>\triangle ABC \sim \triangle F'PF</math>, <math>\frac{AB}{AC}=\frac{F'P}{F'F}</math>, so <math>\frac{425}{510}=\frac{x}{510-d}</math>. Simplifying the LHS and cross-multiplying, we have <math>6x=2550-5d</math>. From the same triangles, we can find that <math>FP=\frac{18}{17}x</math>.
 +
 +
<math>\triangle PEE'</math> is also similar to <math>\triangle F'PF</math>. Since <math>EF'=d</math>, <math>EP=d-x</math>. We now have <math>\frac{PE}{EE'}=\frac{F'P}{FP}</math>, and <math>\frac{d-x}{450-d}=\frac{17}{18}</math>. Cross multiplying, we have <math>18d-18x=450 \cdot 17-17d</math>. Using the previous equation to substitute for <math>x</math>, we have: <cmath>18d-3\cdot2550+15d=450\cdot17-17d</cmath> This is a linear equation in one variable, and we can solve to get <math>d=\boxed{306}</math>
 +
 +
*I did not show the multiplication in the last equation because most of it cancels out when solving.
 +
 +
(Note: I chose <math>F'P</math> to be <math>x</math> only because that is what I had written when originally solving. The solution would work with other choices for <math>x</math>.)
 +
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/FWmrHV1dWPM?t=396
 +
 +
~ pi_is_3.14
  
 
== See also ==
 
== See also ==

Latest revision as of 12:08, 29 July 2024

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); 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.

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

[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 3

[asy] size(200); pathpen = black; pointpen = black + linewidth(0.6); pen s = fontsize(10); // Define points pair C = (0,0), A = (510,0); pair B = IP(circle(C,450),circle(A,425)); // Construct remaining points pair Da = IP(circle(A,289),A--B); pair E = IP(circle(C,324),B--C); pair Ea = IP(circle(B,270),B--C); pair D = IP(Ea--(Ea+A-C),A--B); pair F = IP(Da--(Da+C-B),A--C); pair Fa = IP(E--(E+A-B),A--C); // Draw the main triangle draw(A--B--C--cycle); dot(MP("A",A,s)); dot(MP("B",B,N,s)); dot(MP("C",C,s)); // Mark and draw the other points 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)); // Draw connecting lines draw(D--Ea); draw(Da--F); draw(Fa--E); // Label distances label("450", (B+C)/2, NW); label("425", (A+B)/2, NE); label("510", (A+C)/2, S); // Additional point P 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 ($\triangle ABC \sim \triangle DD'P \sim \triangle PEE' \sim \triangle F'PF$). 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

[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]

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 OmegaLearn

https://youtu.be/FWmrHV1dWPM?t=396

~ pi_is_3.14

See also

1986 AIME (Problems • Answer Key • Resources)
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