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Difference between revisions of "2001 AIME I Problems/Problem 7"

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== Problem ==
 
== Problem ==
−
Triangle <math>ABC</math> has <math>AB=21</math>, <math>AC=22</math> and <math>BC=20</math>. Points <math>D</math> and <math>E</math> are located on <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, such that <math>\overline{DE}</math> is parallel to <math>\overline{BC}</math> and contains the center of the inscribed circle of triangle <math>ABC</math>. Then <math>DE=m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
+
[[Triangle]] <math>ABC</math> has <math>AB=21</math>, <math>AC=22</math> and <math>BC=20</math>. Points <math>D</math> and <math>E</math> are located on <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, such that <math>\overline{DE}</math> is [[parallel]] to <math>\overline{BC}</math> and contains the center of the [[incircle|inscribed circle]] of triangle <math>ABC</math>. Then <math>DE=m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
−
== Solution ==
+
__TOC__
−
By Heron's formula, the area of the whole triangle is <math>21\sqrt{1311}/2</math>. Since the area of a triangle is the inradius times the semiperimeter, the inradius is <math>\sqrt{1311}/6</math>. The ratio of the heights of triangles ADE and ABC is equal to the ratio between sides DE and BC. Thus, we have <math>(21\sqrt{1311}/40-\sqrt{1311}/6)/(21\sqrt{1311}/40)=x/20</math>. Solving for x gives x=<math>860/63</math>, so the answer is <math>923</math>.
+
== Solution 1 ==
 +
 
 +
<center><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></center>
 +
 
 +
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 ==
 +
 
 +
<center><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></center>
 +
 
 +
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}}{4}</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>.
 +
 
 +
 
 +
 
 +
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>.
 +
 
 +
== Solution 3 ([[mass points]]) ==
 +
 
 +
<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(B,C,A,A--B),d); D(anglebisector(C,B,A,A--C),d);
 +
</asy></center>
 +
 
 +
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>.
 +
 
 +
== 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>
 +
 
 +
 
 +
== 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 ==
 
{{AIME box|year=2001|n=I|num-b=6|num-a=8}}
 
{{AIME box|year=2001|n=I|num-b=6|num-a=8}}
 +
 +
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Revision as of 07:31, 1 December 2022

Problem

Triangle $ABC$ has $AB=21$, $AC=22$ and $BC=20$. Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$, respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$. Then $DE=m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

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]

Let $I$ be the incenter of $\triangle ABC$, so that $BI$ and $CI$ are angle bisectors of $\angle ABC$ and $\angle ACB$ respectively. Then, $\angle BID = \angle CBI = \angle DBI,$ so $\triangle BDI$ is isosceles, and similarly $\triangle CEI$ is isosceles. It follows that $DE = DB + EC$, so the perimeter of $\triangle ADE$ is $AD + AE + DE = AB + AC = 43$. Hence, the ratio of the perimeters of $\triangle ADE$ and $\triangle ABC$ is $\frac{43}{63}$, which is the scale factor between the two similar triangles, and thus $DE = \frac{43}{63} \times 20 = \frac{860}{63}$. Thus, $m + n = \boxed{923}$.

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]

The semiperimeter of $ABC$ is $s = \frac{20 + 21 + 22}{2} = \frac{63}{2}$. By Heron's formula, the area of the whole triangle is $A = \sqrt{s(s-a)(s-b)(s-c)} = \frac{21\sqrt{1311}}{4}$. Using the formula $A = rs$, we find that the inradius is $r = \frac{A}{s} = \frac{\sqrt{1311}}6$. Since $\triangle ADE \sim \triangle ABC$, the ratio of the heights of triangles $ADE$ and $ABC$ is equal to the ratio between sides $DE$ and $BC$. From $A=\frac{1}{2}bh$, we find $h_{ABC} = \frac{21\sqrt{1311}}{40}$. Thus, we have

$\frac{h_{ADE}}{h_{ABC}} = \frac{h_{ABC}-r}{h_{ABC}} = \frac{21\sqrt{1311}/40-\sqrt{1311}/6}{21\sqrt{1311}/40}=\frac{DE}{20}.$

Solving for $DE$ gives $DE=\frac{860}{63},$ so the answer is $m+n=\boxed{923}$.


Or we have the area of the triangle as $S$. Using the ratio of heights to ratio of bases of $ADE$ and $ABC$ $\frac {\frac {2S}{20}-\frac {2S}{63}}{\frac {2S}{20}}= \frac {DE}{BC(20)}$ from that it is easy to deduce that $DE=\frac{860}{63}$.

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]

Let $P$ be the incenter; then it is be the intersection of all three angle bisectors. Draw the bisector $AP$ to where it intersects $BC$, and name the intersection $F$.

Using the angle bisector theorem, we know the ratio $BF:CF$ is $21:22$, thus we shall assign a weight of $22$ to point $B$ and a weight of $21$ to point $C$, giving $F$ a weight of $43$. In the same manner, using another bisector, we find that $A$ has a weight of $20$. So, now we know $P$ has a weight of $63$, and the ratio of $FP:PA$ is $20:43$. Therefore, the smaller similar triangle $ADE$ is $43/63$ the height of the original triangle $ABC$. So, $DE$ is $43/63$ the size of $BC$. Multiplying this ratio by the length of $BC$, we find $DE$ is $860/63 = m/n$. Therefore, $m+n=\boxed{923}$.

Solution 4 (Faster)

More directly than Solution 2, we have \[DE=BC\left(\frac{h_a-r}{h_a}\right)=20\left(1-\frac{r}{\frac{[ABC]}{\frac{BC}{2}}}\right)=20\left(1-\frac{10r}{sr}\right)=20\left(1-\frac{10}{\frac{63}{2}}\right)=\frac{860}{63}\implies \boxed{923}.\]


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]

Let the angle bisector of $\angle{A}$ intersects $BC$ at $F$.

Applying the Angle Bisector Theorem on $\angle{A}$ we have \[\frac{AB}{BF}=\frac{AC}{CF}\] \[BF=BC\cdot(\frac{AB}{AB+AC})\] \[BF=\frac{420}{43}\] Since $BP$ is the angle bisector of $\angle{B}$, we can once again apply the Angle Bisector Theorem on $\angle{B}$ which gives \[\frac{BA}{AP}=\frac{BF}{FP}\] \[\frac{AP}{PF}=\frac{AB}{BF}=\frac{41}{20}\] Since $\bigtriangleup ADE\sim\bigtriangleup ABC$ we have \[\frac{DE}{BC}=\frac{AP}{AF}\] \[DE=BC\cdot(\frac{AP}{(\frac{61}{41})\cdot AP})\] Solving gets $DE=\frac{860}{63}$. Thus $m+n=860+63=\boxed{923}$.

~ Nafer

Solution 6

Let $A'$ be the foot of the altitude from $A$ to $\overline {BC}$ and $K$ be the foot of the altitude from $A$ to $\overline{DE}$. Evidently, \[\frac{AK}{AA'} = 1- \frac{r}{AA'} = 1 - \frac{T/s}{T/BC}\] where $r$ is the inradius, $T = [ABC]$, and $s$ is the semiperimeter. So, \[\frac{AK}{AA'} = 1 - \frac{BC}{s} = 1 - \frac{20}{63}= \frac{43}{63}\] Therefore, by similar triangles, we have $DE = BC * \frac{AK}{AA'} = 20 * \frac{AK}{AA'}= \boxed{\frac{860}{63}}$.


Solution 7

Label $P$ the point the angle bisector of $A$ intersects ${BC}$. First we find ${BP}$ and ${PC}$. By the Angle Bisector Theorem, $\frac{BP}{PC} = \frac{21}{22}$ and solving for each using the fact that ${BC} = 20$, we see that ${BP} = \frac{420}{43}$ and $PC = \frac{440}{43}$. 

Because ${AP}$ is the angle bisector of $<A$, we can simply calculate it using Stewarts,

\[{AP} = 21*22 - \frac{440}{43}\cdot\frac{420}{43}\] \[{AP} = 21*22 - \frac{440\cdot420}{43^2}\]

Now we can calculate what ${AO}$ is. Using the formula to find the distance from a vertex to the incenter, ${AO} = \frac{43}{63} \cdot[21\cdot22 - \frac{420*440}{43^2}] = \frac{43^2\cdot22 - 20\cdot440}{43\cdot3}$.

Now because $\triangle{APE} ~ \triangle{ABC}$, we can find ${DE}$ by $\frac{AO}{AP} \cdot 20$. Dividing and simplifying, we see that $\frac{1}{21}\cdot\frac{43}{3}\cdot20 = \frac{860}{63}$. So the answer is $\boxed{923}$

~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

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