1986 AIME Problems/Problem 9
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
Solution
Solution 1 (mass points)
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,
![$d = \frac{2}{\frac{1}{AB} + \frac{1}{BC} + \frac{1}{CA}} = \frac{2}{\frac{1}{510} + \frac{1}{450} + \frac{1}{425}} = \frac{10}{\frac{1}{85}+\frac{1}{90}+\frac{1}{102}}$](http://latex.artofproblemsolving.com/8/1/e/81e13eaae301d4162254ce7c7939261bcbfbe7ad.png)
![$=\frac{10}{\frac{1}{5}\left(\frac{1}{17}+\frac{1}{18}\right)+\frac{1}{102}}=\frac{10}{\frac{1}{5}\cdot\frac{35}{306}+\frac{3}{306}}=\frac{10}{\frac{10}{306}} = \boxed{306}$](http://latex.artofproblemsolving.com/3/f/0/3f0ae95c77dbe02ebb99813f409148054ebe3888.png)
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); 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.
Since is a parallelogram, we find
, and similarly
. So
. Thus
. By the same logic,
.
Since , we have the proportion:
![$\frac{425-d}{425} = \frac{PD}{510} \Longrightarrow PD = 510 - \frac{510}{425}d = 510 - \frac{6}{5}d$](http://latex.artofproblemsolving.com/3/0/8/30821290c12fc0253f1ecd4e539f71adab4964c4.png)
Doing the same with , we find that
. Now,
.
Solution 3
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
answer follows after some hideous computation.
Solution 4
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
.
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 | ||
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