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

m (minor)
(asymptote replacement)
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== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===
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[[Image:1986_AIME-9.png|center]]
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<center><asy>
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size(200);
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pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10);
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pair C=(0,0),A=(510,0),B=IP(circle(C,450),circle(A,425));
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/* construct remaining points */
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pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C);
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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);
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D(MP("A",A,s)--MP("B",B,N,s)--MP("C",C,s)--cycle);
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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));
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D(D--Ea);D(Da--F);D(Fa--E);
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MP("450",(B+C)/2,NW);MP("425",(A+B)/2,NE);MP("510",(A+C)/2);
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</asy></center> <!-- Asymptote replacement for Image:1986_AIME-9.png by azjps -->
  
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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.  
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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.
  
 
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>.  
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Since <math>\triangle DPD' \sim \triangle ABC</math>, we have the [[proportion]]:
 
Since <math>\triangle DPD' \sim \triangle ABC</math>, we have the [[proportion]]:
  
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<cmath>\frac{425-d}{425} = \frac{PD}{510} \Longrightarrow PD = 510 - \frac{510}{425}d = 510 - \frac{6}{5}d</cmath>
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<center><math>\frac{425-d}{425} = \frac{PD}{510} \Longrightarrow PD = 510 - \frac{510}{425}d = 510 - \frac{6}{5}d</math></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 <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>.
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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 <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>
  
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Key theorem: the ratio of the areas of 2 similar triangles is the ratio of a pair of corresponding sides squared.
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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 <math>x</math> and the area of the triangle be <math>A</math>, using the theorem, we get:
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{{AIME box|year=1986|num-b=8|num-a=10}}
 
{{AIME box|year=1986|num-b=8|num-a=10}}
  
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[[Category:Asymptote needed]]
 
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 10:29, 25 April 2008

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); [/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 2

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.

See also

1986 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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