Difference between revisions of "2001 AIME I Problems/Problem 7"
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The [[semiperimeter]] of <math>ABC</math> is <math>s = \frac{20 + 21 + 22}{2} = \frac{63}{2}</math>. By [[Heron's formula]], the area of the whole triangle is <math>A = \sqrt{s(s-a)(s-b)(s-c)} = \frac{21\sqrt{1311}}2</math>. Using the formula <math>A = rs</math>, we find that the [[inradius]] is <math>r = \frac{A}{s} = \frac{\sqrt{1311}}6</math>. Since <math>\triangle ADE \sim \triangle ABC</math>, the ratio of the heights of triangles <math>ADE</math> and <math>ABC</math> is equal to the ratio between sides <math>DE</math> and <math>BC</math>. From <math>A=\frac{1}{2}bh</math>, we find <math>h_{ABC} = \frac{21\sqrt{1311}}{40}</math>. Thus, we have | The [[semiperimeter]] of <math>ABC</math> is <math>s = \frac{20 + 21 + 22}{2} = \frac{63}{2}</math>. By [[Heron's formula]], the area of the whole triangle is <math>A = \sqrt{s(s-a)(s-b)(s-c)} = \frac{21\sqrt{1311}}2</math>. Using the formula <math>A = rs</math>, we find that the [[inradius]] is <math>r = \frac{A}{s} = \frac{\sqrt{1311}}6</math>. Since <math>\triangle ADE \sim \triangle ABC</math>, the ratio of the heights of triangles <math>ADE</math> and <math>ABC</math> is equal to the ratio between sides <math>DE</math> and <math>BC</math>. From <math>A=\frac{1}{2}bh</math>, we find <math>h_{ABC} = \frac{21\sqrt{1311}}{40}</math>. Thus, we have | ||
<center><math>\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}.</math></center> Solving for <math>DE</math> gives <math>DE=\frac{860}{63},</math> so the answer is <math>m+n=\boxed{923}</math>. | <center><math>\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}.</math></center> Solving for <math>DE</math> gives <math>DE=\frac{860}{63},</math> so the answer is <math>m+n=\boxed{923}</math>. | ||
+ | |||
+ | |||
+ | |||
+ | Or we have the area of the triangle as <math>S</math>. | ||
+ | Using the ratio of heights to ratio of bases of <math>ADE</math> and <math>ABC</math> | ||
+ | <math>\frac {\frac {2S}{20}-\frac {2S}{63}}{\frac {2S}{20}}= \frac {DE}{BC(20)}</math> | ||
+ | from that it is easy to deduce that <math>DE=\frac{860}{63}</math> | ||
+ | Same as above, but simpler | ||
=== Solution 2 ([[mass points]]) === | === Solution 2 ([[mass points]]) === |
Revision as of 19:20, 18 February 2010
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
.
Solution
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(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

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
Same as above, but simpler
Solution 2 (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 incircle; 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,
.
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 |