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

Revision as of 20:26, 24 November 2020

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 $DE = BC * \frac{AK}{AA'} = 20 * \frac{AK}{AA'}= \boxed{\frac{860}{63}}$ .

@@ Line 2: / Line 2: @@
 [[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 ===
+== 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);
@@ Line 12: / Line 26: @@
 </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}}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}}{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>.
-=== Solution 2 ===
-Since the point is the center (call it P) of the inscribed circle, it must be the intersection of all three angle bisectors. Drawing the bisector AP, to where it intersects BC, we shall call this 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=923.
+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>.
 == See also ==
@@ Line 22: / Line 96: @@
 [[Category:Intermediate Geometry Problems]]
+{{MAA Notice}}

2001 AIME I (Problems • Answer Key • Resources)
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