1986 AIME Problems/Problem 9: Difference between revisions

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

@@ Line 1: / Line 1: @@
+__TOC__
 == Problem ==
-In <math>\triangle ABC</math>, <math>AB= 425</math>, <math>BC=450</math>, and <math>AC=510</math>. An interior [[point]] <math>P</math> is then drawn, and [[segment]]s are drawn through <math>P</math> [[parallel]] to the sides of the [[triangle]]. If these three segments are of an equal length <math>d</math>, find <math>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 1 ===
+=== 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 <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);
@@ Line 23: / Line 44: @@
 </asy>
-Construct cevians <math>AX</math>, <math>BY</math> and <math>CZ</math> through <math>P</math>.  Place masses of <math>x,y,z</math> on <math>A</math>, <math>B</math> and <math>C</math> respectively; then <math>P</math> has mass <math>x+y+z</math>.
+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 <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 <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><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>
+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 2 ===
+=== Solution 3 ===
 <center><asy>
 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 -->
-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 <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.
-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 <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>.
-Since <math>\triangle DPD' \sim \triangle ABC</math>, we have the [[proportion]]:
+Since <imath>\triangle DPD' \sim \triangle ABC</imath>, we have the [[proportion]]:
-<center><math>\frac{425-d}{425} = \frac{PD}{510} \Longrightarrow PD = 510 - \frac{510}{425}d = 510 - \frac{6}{5}d</math></center>
+<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 <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 <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 3 ===
+=== Solution 4 ===
 Define the points the same as above.
-Let <math>[CE'PF] = a</math>, <math>[E'EP] = b</math>, <math>[BEPD'] = c</math>, <math>[D'PD] = d</math>, <math>[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>
 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 {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>
+<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
-<math>\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>\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
-<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>\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 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 <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>
-<math>=\frac{10}{\frac{10}{306}}=\boxed{306}</math>.
+<imath>=\frac{10}{\frac{10}{306}}=\boxed{306}</imath>.
-=== Solution 5 ===
+=== Solution 6 ===
-Refer to the diagram from Solution 2. 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>.
+<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 <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>.
+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>.
-<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*17-17d</math>. Using the previous equation to substitute for <math>x</math>, we have: <cmath>18d-3*2550+15d=450*17-17d</cmath> This is a linear equation in one variable, and we can solve to get <math>d=\boxed{306}</math>
+<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 <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>.)
+(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 ==

1986 AIME (Problems • Answer Key • Resources)
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