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

Latest revision as of 06:57, 30 September 2024

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  is the angle bisector of , 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

Solution 8 (vectors)

To solve this problem, we can use the fact that, in $\triangle ABC$ , the vector representation of the incenter is $\overrightarrow I = \frac{a\overrightarrow A + b\overrightarrow B + c\overrightarrow C}{a+b+c}$ and that that the vector of the foot of the bisector of $\angle BAC$ on $\overline{BC}$ is $\overrightarrow P = \frac{b\overrightarrow B + c\overrightarrow C}{b+c}$ , where $a=BC,$ $b=AC,$ and $c=AB$ .

Let point $A$ be the origin of the coordinate plane. Then, $\overrightarrow A$ is the zero vector, so we can simplify our expression for $\overrightarrow I$ to $\frac{b\overrightarrow B + c\overrightarrow C}{a+b+c}$ . Now, note that the vector components of $\overrightarrow I$ and $\overrightarrow P$ are the same, but they are multiplied by different scalars. Thus, the ratio of these scalars is the ratio of these vectors' magnitudes. Thus, we have $\frac{|\overrightarrow I|}{|\overrightarrow P|}=\frac{\tfrac1{a+b+c}}{\tfrac1{b+c}}=\frac{b+c}{a+b+c}=\frac{43}{63}$ .

Let $D \in \overline{AB}$ and $E \in \overline{AC}$ . Because $\triangle AIE \sim \triangle APC$ , we have $\frac{AI}{AP}=\frac{AE}{AC}$ . Further, because $\triangle ADE \sim \triangle ABC$ , we have $\frac{AE}{AC}=\frac{DE}{BC}$ . Thus, by transitivity, $\frac{AI}{AP}=\frac{DE}{BC}$ . We know that $\frac{AI}{AP}=\frac{43}{63}$ , so $DE=\frac{AI}{AD}\cdot BC = \frac{43}{63}\cdot 20 = \frac{860}{63}$ .

Thus, our answer is $860+63=\boxed{923}$ .

@@ Line 1: / Line 1: @@
 == 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 [[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__
+== 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
+== Solution 8 (vectors) ==
+To solve this problem, we can use the fact that, in <math>\triangle ABC</math>, the vector representation of the incenter is <math>\overrightarrow I = \frac{a\overrightarrow A + b\overrightarrow B + c\overrightarrow C}{a+b+c}</math> and that that the vector of the foot of the bisector of <math>\angle BAC</math> on <math>\overline{BC}</math> is <math>\overrightarrow P = \frac{b\overrightarrow B + c\overrightarrow C}{b+c}</math>, where <math>a=BC,</math> <math>b=AC,</math> and <math>c=AB</math>.
+Let point <math>A</math> be the origin of the coordinate plane. Then, <math>\overrightarrow A</math> is the zero vector, so we can simplify our expression for <math>\overrightarrow I</math> to <math>\frac{b\overrightarrow B + c\overrightarrow C}{a+b+c}</math>. Now, note that the vector components of <math>\overrightarrow I</math> and <math>\overrightarrow P</math> are the same, but they are multiplied by different scalars. Thus, the ratio of these scalars is the ratio of these vectors' magnitudes. Thus, we have <math>\frac{|\overrightarrow I|}{|\overrightarrow P|}=\frac{\tfrac1{a+b+c}}{\tfrac1{b+c}}=\frac{b+c}{a+b+c}=\frac{43}{63}</math>.
+Let <math>D \in \overline{AB}</math> and <math>E \in \overline{AC}</math>. Because <math>\triangle AIE \sim \triangle APC</math>, we have <math>\frac{AI}{AP}=\frac{AE}{AC}</math>. Further, because <math>\triangle ADE \sim \triangle ABC</math>, we have <math>\frac{AE}{AC}=\frac{DE}{BC}</math>. Thus, by transitivity, <math>\frac{AI}{AP}=\frac{DE}{BC}</math>. We know that <math>\frac{AI}{AP}=\frac{43}{63}</math>, so <math>DE=\frac{AI}{AD}\cdot BC = \frac{43}{63}\cdot 20 = \frac{860}{63}</math>.
+Thus, our answer is <math>860+63=\boxed{923}</math>.
 == See also ==
-* [[2001 AIME I Problems]]
+{{AIME box|year=2001|n=I|num-b=6|num-a=8}}
+[[Category:Intermediate Geometry Problems]]
+{{MAA Notice}}

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