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1986 AIME Problems/Problem 9: Difference between revisions

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__TOC__
== Problem ==
== Problem ==
In <math>\displaystyle \triangle ABC</math>, <math>\displaystyle AB= 425</math>, <math>\displaystyle BC=450</math>, and <math>\displaystyle AC=510</math>. An interior [[point]] <math>\displaystyle P</math> is then drawn, and [[segment]]s are drawn through <math>\displaystyle P</math> [[parallel]] to the sides of the [[triangle]]. If these three segments are of an equal length <math>\displaystyle d</math>, find <math>\displaystyle d</math>.
In <imath>\triangle ABC</imath>, <imath>AB= 425</imath>, <imath>BC=450</imath>, and <imath>AC=510</imath>. An interior [[point]] <imath>P</imath> is then drawn, and [[segment]]s are drawn through <imath>P</imath> [[parallel]] to the sides of the [[triangle]]. If these three segments are of an equal length <imath>d</imath>, find <imath>d</imath>.


__TOC__
== Solution ==
== Solution ==
=== Solution 1 ===
=== Solution 1 ===
[[Image:1986_AIME-9.png]]
<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 <imath>D, D', E, E', F, F'</imath> as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are [[similar triangles|similar]] (<imath>\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF</imath>). The remaining three sections are [[parallelogram]]s.
 
By similar triangles, <imath>BE'=\frac{d}{510}\cdot450=\frac{15}{17}d</imath> and <imath>EC=\frac{d}{425}\cdot450=\frac{18}{17}d</imath>. Since <imath>FD'=BC-EE'</imath>, we have <imath>900-\frac{33}{17}d=d</imath>, so <imath>d=\boxed{306}</imath>.
 
===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 <imath>AX</imath>, <imath>BY</imath> and <imath>CZ</imath> through <imath>P</imath>.  Place masses of <imath>x,y,z</imath> on <imath>A</imath>, <imath>B</imath> and <imath>C</imath> respectively; then <imath>P</imath> has mass <imath>x+y+z</imath>.
 
Notice that <imath>Z</imath> has mass <imath>x+y</imath>.  On the other hand, by similar triangles, <imath>\frac{CP}{CZ} = \frac{d}{AB}</imath>.  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, <center><imath> 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}}</imath><imath>= \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}</imath></center>
 
=== Solution 3 ===
<center><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);


Let the points at which the segments hit the triangle be called <math>D, D', E, E', F, F'</math> as shown above. 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>). This is easy to find using repeated [[alternate interior angles]]. The remaining three sections are [[parallelogram]]s, which is also simple to see by the parallel lines.
// Label distances
label("450", (B+C)/2, NW);
label("425", (A+B)/2, NE);
label("510", (A+C)/2, 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>.
// 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 -->


Since <math>\triangle DPD' \sim \triangle ABC</math>, we have the [[proportion]]:
Let the points at which the segments hit the triangle be called <imath>D, D', E, E', F, F'</imath> as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are [[similar triangles|similar]] (<imath>\triangle ABC \sim \triangle DD'P \sim \triangle PEE' \sim \triangle F'PF</imath>). The remaining three sections are [[parallelogram]]s.


<div style="text-align:center;"><math>\frac{425-d}{425} = \frac{PD}{510}</math><br /><math>PD = 510 - \frac{510}{425}d = 510 - \frac{6}{5}d</math></div>
Since <imath>PDAF'</imath> is a parallelogram, we find <imath>PF' = AD</imath>, and similarly <imath>PE = BD'</imath>. So <imath>d = PF' + PE = AD + BD' = 425 - DD'</imath>. Thus <imath>DD' = 425 - d</imath>. By the same logic, <imath>EE' = 450 - d</imath>.


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 = 306</math>.
Since <imath>\triangle DPD' \sim \triangle ABC</imath>, we have the [[proportion]]:


=== Solution 2 ===
<center><imath>\frac{425-d}{425} = \frac{PD}{510} \Longrightarrow PD = 510 - \frac{510}{425}d = 510 - \frac{6}{5}d</imath></center>
 
Doing the same with <imath>\triangle PEE'</imath>, we find that <imath>PE' =510 - \frac{17}{15}d</imath>. Now, <imath>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}</imath>.
 
=== Solution 4 ===
Define the points the same as above.
Define the points the same as above.


Let <math>[CE'PF] = a</math>, <math>[E'EP] = b</math>, <math>\displaystyle [BEPD'] = c \displaystyle</math>, <math>[D'PD] = d</math>, <math>\displaystyle [DAF'P] = e</math> and <math>[F'D'P] = f</math>
Let <imath>[CE'PF] = a</imath>, <imath>[E'EP] = b</imath>, <imath>[BEPD'] = c</imath>, <imath>[D'PD] = d</imath>, <imath>[DAF'P] = e</imath> and <imath>[F'D'P] = f</imath>


Key theorem: the ratio of the areas of 2 similar triangles is the ratio of a pair of corresponding sides squared.
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 <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 <imath>x</imath> and the area of the triangle be <imath>A</imath>, using the theorem, we get:


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


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 <imath>\triangle ABC \sim \triangle F'PF</imath>; since <imath>ADPF'</imath> and <imath>CFPE'</imath> are parallelograms, it is easy to show that <imath>FF' = AC - x</imath>


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>\displaystyle \frac {f}{A} = \left(\frac {AC - x}{AC}\right)^2 = \left(1 - \frac {x}{AC}\right)^2</math>, by symmetry, we have
<imath>\frac {f}{A} = \left(\frac {AC - x}{AC}\right)^2 = \left(1 - \frac {x}{AC}\right)^2</imath>. By symmetry, we have
<math>\displaystyle \frac {d}{A} = \left(1 - \frac {x}{AB}\right)^2</math> and <math>\frac {b}{A} = \left(1 - \frac {x}{BC}\right)^2</math>
<imath>\frac {d}{A} = \left(1 - \frac {x}{AB}\right)^2</imath> and <imath>\frac {b}{A} = \left(1 - \frac {x}{BC}\right)^2</imath>


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>
<imath>1 + \sum_{cyc}\left(1 - \frac {x}{AB}\right) - \frac {x^2}{AB^2} = 0</imath>
<math>1 + \sum_{cyc}1 - 2*\frac {x}{AB} = 0</math>
<imath>\implies 1 + \sum_{cyc}1 - 2 \cdot \frac {x}{AB} = 0</imath>
<math>\displaystyle \frac {2}{\frac {1}{AB} + \frac {1}{BC} + \frac {1}{CA}} = x</math>
<imath>\implies \frac {2}{\frac {1}{AB} + \frac {1}{BC} + \frac {1}{CA}} = x</imath>
answer follows after some hideous computation
and the answer follows after some hideous computation.
 
===Solution 5===
Refer to the diagram in solution 2; let <imath>a^2=[E'EP]</imath>, <imath>b^2=[D'DP]</imath>, and <imath>c^2=[F'FP]</imath>. Now, note that <imath>[E'BD]</imath>, <imath>[D'DP]</imath>, and <imath>[E'EP]</imath> are similar, so through some similarities we find that <imath>\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</imath>. Similarly, we find that <imath>[D'AF]=(b+c)^2</imath> and <imath>[F'CE]=(c+a)^2</imath>, so <imath>[ABC]=(a+b+c)^2</imath>. Now, again from similarity, it follows that <imath>\frac{d}{510}=\frac{a+b}{a+b+c}</imath>, <imath>\frac{d}{450}=\frac{b+c}{a+b+c}</imath>, and <imath>\frac{d}{425}=\frac{c+a}{a+b+c}</imath>, so adding these together, simplifying, and solving gives <imath>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}}</imath>
<imath>=\frac{10}{\frac{10}{306}}=\boxed{306}</imath>.
 
=== 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 <imath>CE'PF</imath>, <imath>AF'PD</imath>, and <imath>BD'PE</imath> are parallelograms, <imath>\overline{DD'} = 425-d</imath>, <imath>\overline{EE'} = 450-d</imath>, and <imath>\overline{FF'} = 510-d</imath>.
 
Let <imath>F'P = x</imath>. Then, because <imath>\triangle ABC \sim \triangle F'PF</imath>, <imath>\frac{AB}{AC}=\frac{F'P}{F'F}</imath>, so <imath>\frac{425}{510}=\frac{x}{510-d}</imath>. Simplifying the LHS and cross-multiplying, we have <imath>6x=2550-5d</imath>. From the same triangles, we can find that <imath>FP=\frac{18}{17}x</imath>.
 
<imath>\triangle PEE'</imath> is also similar to <imath>\triangle F'PF</imath>. Since <imath>EF'=d</imath>, <imath>EP=d-x</imath>. We now have <imath>\frac{PE}{EE'}=\frac{F'P}{FP}</imath>, and <imath>\frac{d-x}{450-d}=\frac{17}{18}</imath>. Cross multiplying, we have <imath>18d-18x=450 \cdot 17-17d</imath>. Using the previous equation to substitute for <imath>x</imath>, 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 <imath>d=\boxed{306}</imath>
 
*I did not show the multiplication in the last equation because most of it cancels out when solving.
 
(Note: I chose <imath>F'P</imath> to be <imath>x</imath> only because that is what I had written when originally solving. The solution would work with other choices for <imath>x</imath>.)
 
== Video Solution by Pi Academy ==
https://youtu.be/fHFD0TEfBnA?si=DRKVU_As7Rv0ou5D
 
~ Pi Academy


== See also ==
== See also ==
{{AIME box|year=1986|num-b=8|num-a=10}}
{{AIME box|year=1986|num-b=8|num-a=10}}
* [[AIME Problems and Solutions]]
* [[American Invitational Mathematics Examination]]
* [[Mathematics competition resources]]


[[Category:Intermediate Geometry Problems]]
[[Category:Intermediate Geometry Problems]]
{{MAA Notice}}

Latest revision as of 14:12, 11 November 2025

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 Pi Academy

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

~ Pi Academy

See also

1986 AIME (Problems • Answer Key • Resources)
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