Difference between revisions of "2001 AIME I Problems/Problem 7"
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− | Let <math>I</math> be the [[incenter]] of <math>\triangle ABC</math>, so that <math>BI</math> and <math>CI</math> are [[angle bisector]]s of <math>\angle ABC</math> and <math>\angle ACB</math> respectively. Then, <math>\angle BID = \angle CBI = \angle DBI,</math> so <math>\triangle BDI</math> is [[Isosceles triangle|isosceles]], and similarly <math>\triangle CEI</math> is isosceles. It follows that <math>DE = DB + EC</math>, so the perimeter of <math>\triangle ADE</math> is <math>AD + AE + DE = AB + AC = 43</math>. Hence, the ratio of the perimeters of <math>\triangle ADE</math> and <math>\triangle ABC</math> is <math>\frac{43}{63}</math>, which is the scale factor between the two similar triangles, and thus <math>DE = \frac{43}{63} \times 20 = \frac{860}{63}</math>. Thus, <math>m + n = \boxed{923}</math>. | + | Let <math>I</math> be the [[incenter]] of <math>\triangle ABC</math>, so that <math>BI</math> and <math>CI</math> are [[angle bisector]]s of <math>\angle ABC</math> and <math>\angle ACB</math> respectively. Then, <math>\angle BID = \angle CBI = \angle DBI,</math> so <math>\triangle BDI</math> is [[Isosceles triangle|isosceles]], and similarly <math>\triangle CEI</math> is isosceles. It follows that <math>DE = DB + EC</math>, so the perimeter of <math>\triangle ADE</math> is <math>AD + AE + DE = AB + AC = 43</math>. Hence, the ratio of the perimeters of <math>\triangle ADE</math> and <math>\triangle ABC</math> is <math>\frac{43}{63}</math>, which is the scale factor between the two similar triangles, and thus <math>DE = \frac{43}{63} \times 20 = \frac{860}{63}</math>. Thus, <math>m + n = \boxed{923}</math>. |
== Solution 2 == | == Solution 2 == | ||
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</asy></center> | </asy></center> | ||
− | Let <math>P</math> be the [[ | + | Let <math>P</math> be the [[incenter]]; then it is be the intersection of all three [[angle bisector]]s. Draw the bisector <math>AP</math> to where it intersects <math>BC</math>, and name the intersection <math>F</math>. |
Using the [[angle bisector theorem]], we know the ratio <math>BF:CF</math> is <math>21:22</math>, thus we shall assign a weight of <math>22</math> to point <math>B</math> and a weight of <math>21</math> to point <math>C</math>, giving <math>F</math> a weight of <math>43</math>. In the same manner, using another bisector, we find that <math>A</math> has a weight of <math>20</math>. So, now we know <math>P</math> has a weight of <math>63</math>, and the ratio of <math>FP:PA</math> is <math>20:43</math>. Therefore, the smaller similar triangle <math>ADE</math> is <math>43/63</math> the height of the original triangle <math>ABC</math>. So, <math>DE</math> is <math>43/63</math> the size of <math>BC</math>. Multiplying this ratio by the length of <math>BC</math>, we find <math>DE</math> is <math>860/63 = m/n</math>. Therefore, <math>m+n=\boxed{923}</math>. | Using the [[angle bisector theorem]], we know the ratio <math>BF:CF</math> is <math>21:22</math>, thus we shall assign a weight of <math>22</math> to point <math>B</math> and a weight of <math>21</math> to point <math>C</math>, giving <math>F</math> a weight of <math>43</math>. In the same manner, using another bisector, we find that <math>A</math> has a weight of <math>20</math>. So, now we know <math>P</math> has a weight of <math>63</math>, and the ratio of <math>FP:PA</math> is <math>20:43</math>. Therefore, the smaller similar triangle <math>ADE</math> is <math>43/63</math> the height of the original triangle <math>ABC</math>. So, <math>DE</math> is <math>43/63</math> the size of <math>BC</math>. Multiplying this ratio by the length of <math>BC</math>, we find <math>DE</math> is <math>860/63 = m/n</math>. Therefore, <math>m+n=\boxed{923}</math>. | ||
− | == Solution 4 ( | + | == Solution 4 (Faster) == |
More directly than Solution 2, we have <cmath>DE=BC\left(\frac{h_a-r}{h_a}\right)=20\left(1-\frac{r}{\frac{[ABC]}{\frac{BC}{2}}}\right)=20\left(1-\frac{10r}{sr}\right)=20\left(1-\frac{10}{\frac{63}{2}}\right)=\frac{860}{63}\implies \boxed{923}.</cmath> | More directly than Solution 2, we have <cmath>DE=BC\left(\frac{h_a-r}{h_a}\right)=20\left(1-\frac{r}{\frac{[ABC]}{\frac{BC}{2}}}\right)=20\left(1-\frac{10r}{sr}\right)=20\left(1-\frac{10}{\frac{63}{2}}\right)=\frac{860}{63}\implies \boxed{923}.</cmath> | ||
+ | |||
+ | |||
+ | == Solution 5 == | ||
+ | |||
+ | Diagram borrowed from Solution 3. | ||
+ | |||
+ | <center><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></center> | ||
+ | |||
+ | Let the angle bisector of <math>\angle{A}</math> intersects <math>BC</math> at <math>F</math>. | ||
+ | |||
+ | Applying the [[Angle Bisector Theorem]] on <math>\angle{A}</math> we have | ||
+ | <cmath>\frac{AB}{BF}=\frac{AC}{CF}</cmath> | ||
+ | <cmath>BF=BC\cdot(\frac{AB}{AB+AC})</cmath> | ||
+ | <cmath>BF=\frac{420}{43}</cmath> | ||
+ | Since <math>BP</math> is the angle bisector of <math>\angle{B}</math>, we can once again apply the Angle Bisector Theorem on <math>\angle{B}</math> which gives | ||
+ | <cmath>\frac{BA}{AP}=\frac{BF}{FP}</cmath> | ||
+ | <cmath>\frac{AP}{PF}=\frac{AB}{BF}=\frac{41}{20}</cmath> | ||
+ | Since <math>\bigtriangleup ADE\sim\bigtriangleup ABC</math> we have | ||
+ | <cmath>\frac{DE}{BC}=\frac{AP}{AF}</cmath> | ||
+ | <cmath>DE=BC\cdot(\frac{AP}{(\frac{61}{41})\cdot AP})</cmath> | ||
+ | Solving gets <math>DE=\frac{860}{63}</math>. Thus <math>m+n=860+63=\boxed{923}</math>. | ||
+ | |||
+ | ~ Nafer | ||
+ | |||
+ | == 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'} = 20 * \frac{AK}{AA'}= \boxed{\frac{860}{63}}</math>. | ||
+ | |||
+ | |||
+ | == Solution 7 == | ||
+ | Label <math>P</math> the point the angle bisector of <math>A</math> intersects <math>{BC}</math>. First we find <math>{BP}</math> and <math>{PC}</math>. By the Angle Bisector Theorem, <math>\frac{BP}{PC} = \frac{21}{22}</math> and solving for each using the fact that <math>{BC} = 20</math>, we see that <math>{BP} = \frac{420}{43}</math> and <math>PC = \frac{440}{43}</math>. | ||
+ | |||
+ | Because <math>{AP}</math> is the angle bisector of <math><A</math>, we can simply calculate it using Stewarts, | ||
+ | <cmath> {AP} = 21*22 - \frac{440}{43}\cdot\frac{420}{43}</cmath> | ||
+ | <cmath> {AP} = 21*22 - \frac{440\cdot420}{43^2}</cmath> | ||
+ | |||
+ | Now we can calculate what <math>{AO}</math> is. Using the formula to find the distance from a vertex to the incenter, <math>{AO} = \frac{43}{63} \cdot[21\cdot22 - \frac{420*440}{43^2}] = \frac{43^2\cdot22 - 20\cdot440}{43\cdot3}</math>. | ||
+ | |||
+ | Now because <math>\triangle{APE} ~ \triangle{ABC}</math>, we can find <math>{DE}</math> by <math>\frac{AO}{AP} \cdot 20</math>. Dividing and simplifying, we see that <math>\frac{1}{21}\cdot\frac{43}{3}\cdot20 = \frac{860}{63}</math>. So the answer is <math>\boxed{923}</math> | ||
+ | |||
+ | ~YBSuburbanTea | ||
== See also == | == See also == |
Latest revision as of 08:31, 1 December 2022
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
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
.
Solution 7
Label the point the angle bisector of
intersects
. First we find
and
. By the Angle Bisector Theorem,
and solving for each using the fact that
, we see that
and
.
Becauseis the angle bisector of
, we can simply calculate it using Stewarts,
Now we can calculate what is. Using the formula to find the distance from a vertex to the incenter,
.
Now because , we can find
by
. Dividing and simplifying, we see that
. So the answer is
~YBSuburbanTea
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.