Difference between revisions of "2001 AIME I Problems/Problem 7"
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+ | == Solution 6 == | ||
+ | Let <math>A'</math> be the foot of the altitude from <math>A</math> to <math>\overline {BC}</math> and <math>K</math> be the foot of the altitude from <math>A</math> to <math>\overline{DE}</math>. Evidently, <cmath>\frac{AK}{AA'} = 1- \frac{r}{AA'} = 1 - \frac{T/s}{T/BC}</cmath> where <math>r</math> is the inradius, <math>T = [ABC]</math>, and <math>s</math> is the semiperimeter. So, <cmath>\frac{AK}{AA'} = 1 - \frac{BC}{s} = 1 - \frac{20}{63}= \frac{43}{63}</cmath> Therefore, by similar triangles, we have <math>DE = BC * \frac{AK}{AA'} = \boxed{\frac{860}{63}}</math>. | ||
== See also == | == See also == |
Revision as of 19:25, 24 November 2020
Problem
Triangle has
,
and
. Points
and
are located on
and
, respectively, such that
is parallel to
and contains the center of the inscribed circle of triangle
. Then
, where
and
are relatively prime positive integers. Find
.
Contents
[hide]Solution 1
![[asy] pointpen = black; pathpen = black+linewidth(0.7); pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C); D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle); D(MP("I",I,NE)); D(MP("E",E,NE)--MP("D",D,NW)); // D((A.x,0)--A,linetype("4 4")+linewidth(0.7)); D((I.x,0)--I,linetype("4 4")+linewidth(0.7)); D(rightanglemark(B,(A.x,0),A,30)); D(B--I--C); MP("20",(B+C)/2); MP("21",(A+B)/2,NW); MP("22",(A+C)/2,NE); [/asy]](http://latex.artofproblemsolving.com/a/e/4/ae47e3711da65c309f6b7dc3c0e55287b8c8a5fc.png)
Let be the incenter of
, so that
and
are angle bisectors of
and
respectively. Then,
so
is isosceles, and similarly
is isosceles. It follows that
, so the perimeter of
is
. Hence, the ratio of the perimeters of
and
is
, which is the scale factor between the two similar triangles, and thus
. Thus,
.
Solution 2
![[asy] pointpen = black; pathpen = black+linewidth(0.7); pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C); D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle); D(incircle(A,B,C)); D(MP("I",I,NE)); D(MP("E",E,NE)--MP("D",D,NW)); D((A.x,0)--A,linetype("4 4")+linewidth(0.7)); D((I.x,0)--I,linetype("4 4")+linewidth(0.7)); D(rightanglemark(B,(A.x,0),A,30)); MP("20",(B+C)/2); MP("21",(A+B)/2,NW); MP("22",(A+C)/2,NE); [/asy]](http://latex.artofproblemsolving.com/3/0/8/308015d62aeb9917bd2a032c366423e1e0529c1f.png)
The semiperimeter of is
. By Heron's formula, the area of the whole triangle is
. Using the formula
, we find that the inradius is
. Since
, the ratio of the heights of triangles
and
is equal to the ratio between sides
and
. From
, we find
. Thus, we have
![$\frac{h_{ADE}}{h_{ABC}} = \frac{h_{ABC}-r}{h_{ABC}} = \frac{21\sqrt{1311}/40-\sqrt{1311}/6}{21\sqrt{1311}/40}=\frac{DE}{20}.$](http://latex.artofproblemsolving.com/c/7/a/c7ada10f62e3db0133421cd545a87b5575997be9.png)
Solving for gives
so the answer is
.
Or we have the area of the triangle as .
Using the ratio of heights to ratio of bases of
and
from that it is easy to deduce that
.
Solution 3 (mass points)
![[asy] pointpen = black; pathpen = black+linewidth(0.7); pen d = linewidth(0.7) + linetype("4 4"); pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C); D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle); D(incircle(A,B,C)); D(MP("P",I,(1,2))); D(MP("E",E,NE)--MP("D",D,NW)); MP("20",(B+C)/2); MP("21",(A+B)/2,NW); MP("22",(A+C)/2,NE); /* construct angle bisectors */ path anglebisector (pair X, pair Y, pair Z, path K) { return Y -- IP(Y -- Y + 30 * (bisectorpoint(X,Y,Z)-Y) , K); } D(anglebisector(C,A,B,B--C), d); D(anglebisector(B,C,A,A--B),d); D(anglebisector(C,B,A,A--C),d); [/asy]](http://latex.artofproblemsolving.com/7/c/7/7c752cc350ded25b2c947d336d77bc6ccf240f96.png)
Let be the incenter; then it is be the intersection of all three angle bisectors. Draw the bisector
to where it intersects
, and name the intersection
.
Using the angle bisector theorem, we know the ratio is
, thus we shall assign a weight of
to point
and a weight of
to point
, giving
a weight of
. In the same manner, using another bisector, we find that
has a weight of
. So, now we know
has a weight of
, and the ratio of
is
. Therefore, the smaller similar triangle
is
the height of the original triangle
. So,
is
the size of
. Multiplying this ratio by the length of
, we find
is
. Therefore,
.
Solution 4 (Faster)
More directly than Solution 2, we have
Solution 5
Diagram borrowed from Solution 3.
![[asy] pointpen = black; pathpen = black+linewidth(0.7); pen d = linewidth(0.7) + linetype("4 4"); pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C); D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle); D(incircle(A,B,C)); D(MP("P",I,(1,2))); D(MP("E",E,NE)--MP("D",D,NW)); MP("20",(B+C)/2); MP("21",(A+B)/2,NW); MP("22",(A+C)/2,NE); /* construct angle bisectors */ path anglebisector (pair X, pair Y, pair Z, path K) { return Y -- IP(Y -- Y + 30 * (bisectorpoint(X,Y,Z)-Y) , K); } D(anglebisector(C,A,B,B--C), d); D(anglebisector(C,B,A,A--C),d); [/asy]](http://latex.artofproblemsolving.com/8/a/7/8a737cd3a7bc5d89bf1865f77a48dfdfe147b5d9.png)
Let the angle bisector of intersects
at
.
Applying the Angle Bisector Theorem on we have
Since
is the angle bisector of
, we can once again apply the Angle Bisector Theorem on
which gives
Since
we have
Solving gets
. Thus
.
~ Nafer
Solution 6
Let be the foot of the altitude from
to
and
be the foot of the altitude from
to
. Evidently,
where
is the inradius,
, and
is the semiperimeter. So,
Therefore, by similar triangles, we have
.
See also
2001 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.