Difference between revisions of "1986 AIME Problems/Problem 9"
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__TOC__ | __TOC__ | ||
== Solution == | == 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 <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, < | + | 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 | + | === 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) | + | pointpen = black + linewidth(0.6); |
− | / | + | pen s = fontsize(10); |
− | pair Da=IP( | + | |
− | pair D=IP(Ea--(Ea+A-C),A--B) | + | // Define points |
− | + | 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)); |
− | + | ||
− | + | // Construct remaining points | |
− | / | + | 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 | + | 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 | + | === 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 { | + | <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 | |
− | <math>\frac {f + d + b}{A} + 1 = x^2 | + | <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> | + | 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> | + | <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 | + | <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 | + | ===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 Pi Academy == | ||
+ | https://youtu.be/fHFD0TEfBnA?si=DRKVU_As7Rv0ou5D | ||
+ | |||
+ | ~ Pi Academy | ||
== See also == | == See also == |
Latest revision as of 21:04, 8 December 2024
Problem
In ,
,
, and
. An interior point
is then drawn, and segments are drawn through
parallel to the sides of the triangle. If these three segments are of an equal length
, find
.
Contents
[hide]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]](http://latex.artofproblemsolving.com/8/e/1/8e19916c4ed37209e832673ba0274846bb4c4246.png)
Let the points at which the segments hit the triangle be called as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are similar (
). The remaining three sections are parallelograms.
By similar triangles, and
. Since
, we have
, so
.
Solution 2
Construct cevians ,
and
through
. Place masses of
on
,
and
respectively; then
has mass
.
Notice that has mass
. On the other hand, by similar triangles,
. Hence by mass points we find that
Similarly, we obtain
Summing these three equations yields
Hence,


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]](http://latex.artofproblemsolving.com/0/8/a/08ae6b7affdbd9f7a37f66ddf918895508ad791c.png)
Let the points at which the segments hit the triangle be called as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are similar (
). The remaining three sections are parallelograms.
Since is a parallelogram, we find
, and similarly
. So
. Thus
. By the same logic,
.
Since , we have the proportion:

Doing the same with , we find that
. Now,
.
Solution 4
Define the points the same as above.
Let ,
,
,
,
and
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 and the area of the triangle be
, using the theorem, we get:
,
,
.
Adding all these together and using
we get
Using corresponding angles from parallel lines, it is easy to show that ; since
and
are parallelograms, it is easy to show that
Now we have the side length ratio, so we have the area ratio
. By symmetry, we have
and
Substituting these into our initial equation, we have
and the answer follows after some hideous computation.
Solution 5
Refer to the diagram in solution 2; let ,
, and
. Now, note that
,
, and
are similar, so through some similarities we find that
. Similarly, we find that
and
, so
. Now, again from similarity, it follows that
,
, and
, so adding these together, simplifying, and solving gives
.
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]](http://latex.artofproblemsolving.com/8/e/1/8e19916c4ed37209e832673ba0274846bb4c4246.png)
Refer to the diagram above. Notice that because ,
, and
are parallelograms,
,
, and
.
Let . Then, because
,
, so
. Simplifying the LHS and cross-multiplying, we have
. From the same triangles, we can find that
.
is also similar to
. Since
,
. We now have
, and
. Cross multiplying, we have
. Using the previous equation to substitute for
, we have:
This is a linear equation in one variable, and we can solve to get
- I did not show the multiplication in the last equation because most of it cancels out when solving.
(Note: I chose to be
only because that is what I had written when originally solving. The solution would work with other choices for
.)
Video Solution by Pi Academy
https://youtu.be/fHFD0TEfBnA?si=DRKVU_As7Rv0ou5D
~ Pi Academy
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.