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

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== Solution 6 ==
 
== 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>.
 
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>.
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== Solution 7 ==
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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>.
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Because <math>{AP}</math> is the angle bisector of <math><A</math>, we can simply calculate it using Stewarts,
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<cmath> {AP} = 21*22 - \frac{440}{43}\cdot\frac{420}{43}</cmath>
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<cmath> {AP} = 21*22 - \frac{440\cdot420}{43^2}</cmath>
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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>.
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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>
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~YBSuburbanTea
  
 
== See also ==
 
== See also ==

Revision as of 07:30, 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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