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Difference between revisions of "2003 AIME I Problems/Problem 15"

(solution (mass points) by me@home, Asymptote [myself])
(→‎Solution 4 (Overpowered Projective Geometry!!))
 
(16 intermediate revisions by 10 users not shown)
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In <math> \triangle ABC, AB = 360, BC = 507, </math> and <math> CA = 780. </math> Let <math> M </math> be the [[midpoint]] of <math> \overline{CA}, </math> and let <math> D </math> be the point on <math> \overline{CA} </math> such that <math> \overline{BD} </math> bisects angle <math> ABC. </math> Let <math> F </math> be the point on <math> \overline{BC} </math> such that <math> \overline{DF} \perp \overline{BD}. </math> Suppose that <math> \overline{DF} </math> meets <math> \overline{BM} </math> at <math> E. </math> The ratio <math> DE: EF </math> can be written in the form <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers. Find <math> m + n. </math>
 
In <math> \triangle ABC, AB = 360, BC = 507, </math> and <math> CA = 780. </math> Let <math> M </math> be the [[midpoint]] of <math> \overline{CA}, </math> and let <math> D </math> be the point on <math> \overline{CA} </math> such that <math> \overline{BD} </math> bisects angle <math> ABC. </math> Let <math> F </math> be the point on <math> \overline{BC} </math> such that <math> \overline{DF} \perp \overline{BD}. </math> Suppose that <math> \overline{DF} </math> meets <math> \overline{BM} </math> at <math> E. </math> The ratio <math> DE: EF </math> can be written in the form <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers. Find <math> m + n. </math>
  
−
== Solution ==
+
== Solution 1 ==
 +
 
 +
In the following, let the name of a point represent the mass located there. Since we are looking for a ratio, we assume that <math>AB=120</math>, <math>BC=169</math>, and <math>CA=260</math> in order to simplify our computations.
 +
 
 +
First, reflect point <math>F</math> over angle bisector <math>BD</math> to a point <math>F'</math>.
 
<center><asy>
 
<center><asy>
 
size(400); pointpen = black; pathpen = black+linewidth(0.7);  
 
size(400); pointpen = black; pathpen = black+linewidth(0.7);  
−
pair A=(0,0),C=(7.8,0),B=IP(CR(A,3.6),CR(C,5.07)), M=(A+C)/2, Da = bisectorpoint(A,B,C), D=IP(B--B+(Da-B)*10,A--C), F=IP(D--D+10*(B-D)*dir(270),B--C), E=IP(B--M,D--F); /* scale down by 100x */
+
pair A=(0,0),C=(7.8,0),B=IP(CR(A,3.6),CR(C,5.07)), M=(A+C)/2, Da = bisectorpoint(A,B,C), D=IP(B--B+(Da-B)*10,A--C), F=IP(D--D+10*(B-D)*dir(270),B--C), E=IP(B--M,D--F);pair Fprime=2*D-F; /* scale down by 100x */
−
D(MP("A",A)--MP("B",B,N)--MP("C",C)--cycle); D(B--D(MP("D",D))--D(MP("F",F,NE))); D(B--D(MP("M",M))); MP("E",E,NE); D(rightanglemark(F,D,B,4)); MP("390",(M+C)/2); MP("390",(M+C)/2); MP("360",(A+B)/2,NW); MP("507",(B+C)/2,NE);
+
D(MP("A",A,NW)--MP("B",B,N)--MP("C",C)--cycle); D(B--D(MP("D",D))--D(MP("F",F,NE))); D(B--D(MP("M",M)));D(A--MP("F'",Fprime,SW)--D); MP("E",E,NE); D(rightanglemark(F,D,B,4)); MP("390",(M+C)/2); MP("390",(M+C)/2); MP("360",(A+B)/2,NW); MP("507",(B+C)/2,NE);
 
</asy></center>
 
</asy></center>
 +
As <math>BD</math> is an angle bisector of both triangles <math>BAC</math> and <math>BF'F</math>, we know that <math>F'</math> lies on <math>AB</math>. We can now balance triangle <math>BF'C</math> at point <math>D</math> using mass points.
  
−
For computation, instead consider the triangle as above except <math>AB = 120,BC = 169,CA = 260</math>. In the following, let the name of a point represent the mass located there.
+
By the [[Angle Bisector Theorem]], we can place [[mass points]] on <math>C,D,A</math> of <math>120,\,289,\,169</math> respectively. Thus, a mass of <math>\frac {289}{2}</math> belongs at both <math>F</math> and <math>F'</math> because BD is a median of triangle <math>BF'F</math> . Therefore, <math>CB/FB=\frac{289}{240}</math>.
  
−
By the [[Angle Bisector Theorem]], we can place [[mass points]] on <math>C,D,A</math> of <math>120,\,289,\,169</math> respectively. Thus, a mass of <math>\frac {289}{2}</math> belongs at <math>F</math> (seen by reflecting <math>F</math> across <math>BD</math>, to an image which lies on <math>AB</math>).
+
Now, we reassign mass points to determine <math>FE/FD</math>. This setup involves <math>\triangle CFD</math> and [[transversal]] <math>MEB</math>. For simplicity, put masses of <math>240</math> and <math>289</math> at <math>C</math> and <math>F</math> respectively. To find the mass we should put at <math>D</math>, we compute <math>CM/MD</math>. Applying the Angle Bisector Theorem again and using the fact <math>M</math> is a midpoint of <math>AC</math>, we find
−
Having determined <math>CB/CF</math>, we reassign mass points to determine <math>FE/FD</math>. This setup involves <math>\tri CFD</math> and [[transversal]] <math>MEB</math>. For simplicity, put masses of <math>240,289</math> at <math>C,F</math>. To find the mass we should put at <math>D</math>, we compute <math>CM/MD</math>: applying the Angle Bisector Theorem again and using the fact <math>M</math> is a midpoint, we find
 
 
<cmath>
 
<cmath>
−
\frac {CM}{MD} = \frac {169\cdot\frac {260}{289} - 130}{130} = \frac {49}{289}
+
\frac {MD}{CM} = \frac {\frac{169}{289}\cdot 260 - 130}{130} = \frac {49}{289}
 
</cmath>
 
</cmath>
 
At this point we could find the mass at <math>D</math> but it's unnecessary.
 
At this point we could find the mass at <math>D</math> but it's unnecessary.
 
<cmath>
 
<cmath>
−
\frac {DE}{EF} = \frac {F}{D} = \frac {F}{C}\frac {C}{D} = \frac {289}{240}\frac {49}{289} = \boxed{\frac {49}{240}}
+
\frac {DE}{EF} = \frac {F}{D} = \frac {F}{C}\cdot\frac {C}{D} = \frac {289}{240}\cdot\frac {49}{289} = \boxed{\frac {49}{240}}
 
</cmath>
 
</cmath>
 
and the answer is <math>49 + 240 = \boxed{289}</math>.
 
and the answer is <math>49 + 240 = \boxed{289}</math>.
 +
 +
== Solution 2 ==
 +
By the Angle Bisector Theorem, we know that <math>[CBD]=\frac{169}{289}[ABC]</math>. Therefore, by finding the area of triangle <math>CBD</math>, we see that <cmath>\frac{507\cdot BD}{2}\sin\frac{B}{2}=\frac{169}{289}[ABC].</cmath>
 +
Solving for <math>BD</math> yields <cmath>BD=\frac{2[ABC]}{3\cdot289\sin\frac{B}{2}}.</cmath>
 +
Furthermore, <math>\cos\frac{B}{2}=\frac{BD}{BF}</math>, so <cmath>BF=\frac{BD}{\cos\frac{B}{2}}=\frac{2[ABC]}{3\cdot289\sin\frac{B}{2}\cos\frac{B}{2}}.</cmath>
 +
Now by the identity <math>2\sin\frac{B}{2}\cos\frac{B}{2}=\sin B</math>, we get <cmath>BF=\frac{4[ABC]}{3\cdot289\sin B}.</cmath>
 +
But then <math>[ABC]=\frac{360\cdot 507}{2}\sin B</math>, so <math>BF=\frac{240}{289}\cdot 507</math>. Thus <math>BF:FC=240:49</math>.
 +
 +
Now by the Angle Bisector Theorem, <math>CD=\frac{169}{289}\cdot 780</math>, and we know that <math>MC=\frac{1}{2}\cdot 780</math> so <math>DM:MC=\frac{169}{289}-\frac{1}{2}:\frac{1}{2}=49:289</math>.
 +
 +
We can now use mass points on triangle CBD. Assign a mass of <math>240\cdot 49</math> to point <math>C</math>. Then <math>D</math> must have mass <math>240\cdot 289</math> and <math>B</math> must have mass <math>49\cdot 49</math>. This gives <math>F</math> a mass of <math>240\cdot 49+49\cdot 49=289\cdot 49</math>. Therefore, <math>DE:EF=\frac{289\cdot 49}{240\cdot 289}=\frac{49}{240}</math>, giving us an answer of <math>\boxed{289}.</math>
 +
 +
== Solution 3 ==
 +
Let <math>\angle{DBM}=\theta</math> and <math>\angle{DBC}=\alpha</math>. Then because <math>BM</math> is a median we have <math>360\sin{(\alpha+\theta)}=507\sin{(\alpha-\theta)}</math>. Now we know <cmath>\sin{(\alpha+\theta)}=\sin{\alpha}\cos{\theta}+\sin{\theta}\cos{\alpha}=\dfrac{DF\cdot BD}{BF\cdot BE}+\dfrac{DE\cdot BD}{BE\cdot BF}=\dfrac{BD(DF+DE)}{BF\cdot BE}</cmath> Expressing the area of <math>\triangle{BEF}</math> in two ways we have <cmath>\dfrac{1}{2}BE\cdot BF\sin{(\alpha-\theta)}=\dfrac{1}{2}EF\cdot BD</cmath> so <cmath>\sin{(\alpha-\theta)}=\dfrac{EF\cdot BD}{BF\cdot BE}</cmath> Plugging this in we have <cmath>\dfrac{360\cdot BD(DF+DE)}{BF\cdot BE}=\dfrac{507\cdot BD\cdot EF}{BF\cdot BE}</cmath> so <math>\dfrac{DF+DE}{EF}=\dfrac{507}{360}</math>. But <math>DF=DE+EF</math>, so this simplifies to <math>1+\dfrac{2DE}{EF}=\dfrac{507}{360}=\dfrac{169}{120}</math>, and thus <math>\dfrac{DE}{EF}=\dfrac{49}{240}</math>, and <math>m+n=\boxed{289}</math>.
 +
 +
== Solution 4 (Overpowered Projective Geometry!!) ==
 +
Firstly, angle bisector theorem yields <math>\frac{CD}{AD} = \frac{507}{360} = \frac{169}{120}</math>. We're given that <math>AM=MC</math>. Therefore, the cross ratio
 +
 +
<cmath>
 +
(A,C;M,D) = \frac{AM(CD)}{AD(MC)} = \frac{169}{120}
 +
</cmath>
 +
 +
We need a fourth point for this cross ratio to be useful, so reflect point <math>F</math> over angle bisector <math>BD</math> to a point <math>F'</math>. Then <math>\triangle BFF'</math> is isosceles and <math>BD</math> is an altitude so <math>DF = DF'</math>. Therefore,
 +
 +
<cmath>
 +
(A,C;M,D) = (F,F';D,E) \implies \frac{FD(EF')}{EF(DF')} = \frac{EF'}{EF} = \frac{169}{120}
 +
</cmath>
 +
 +
All that's left is to fiddle around with the ratios:
 +
 +
<cmath>
 +
\frac{EF'}{EF} = \frac{ED+DF'}{EF} = \frac{EF+2DE}{EF} = 1\ +\ 2\left(\frac{DE}{EF}\right) \implies \frac{DE}{EF} = \frac{49}{240} \implies \boxed{289}
 +
</cmath>
 +
 +
== Solution 5 (Menelaus + Mass Points) ==
 +
 +
Extend <math>DF</math> to intersect with the extension of <math>AB</math> at <math>G</math>. Notice that <math>\triangle{BDF} \cong \triangle{BDG}</math>, so <math>GD=DF</math>. We now use Menelaus on <math>\triangle{GBF}</math>, as <math>A</math>, <math>D</math>, and <math>C</math> are collinear; this gives us <math>\frac{GA}{BA} \cdot \frac{BC}{FC} \cdot \frac{DF}{GD}=1</math>. As <math>GD=DF</math>, we have <math>\frac{GA}{AB}=\frac{FC}{BC}</math>, hence <math>\frac{GA}{120}=\frac{FC}{169}</math>. Reflect <math>G</math> over <math>A</math> to <math>G'</math>. Note that <math>\frac{G'A}{BA}=\frac{FC}{BC}</math>, and reflexivity, hence <math>\triangle{ABC} \sim \triangle{BG'F}</math>. It's easily concluded from this that <math>G'F \parallel AC</math>, hence <math>G'F \parallel AD</math>. As <math>GD=DF</math>, we have <math>AD</math> is a midsegment of <math>\triangle{GG'F}</math>, thus <math>G'F = 2AD</math>. We now focus on the ratio <math>\frac{BF}{BC}</math>. From similarity, we have <math>\frac{BF}{BC}=\frac{G'F}{AC}=\frac{2AD}{AC}</math>. By the angle bisector theorem, we have <math>AD:DC=120:169</math>, hence <math>AD:AC=120:289</math>, so <math>\frac{BF}{BC}=\frac{240}{289}</math>. We now work out the ratio <math>\frac{DM}{MC}</math>. <math>\frac{DM}{MC}=\frac{CD-MC}{MC}=\frac{CD}{MC}-1=\frac{2CD}{AC}-1=\frac{338}{289}-1=\frac{49}{289}</math>. We now use mass points on <math>\triangle{BDC}</math>. We let the mass of <math>C</math> be <math>240\cdot 49</math>, so the mass of <math>B</math> is <math>49 \cdot 49</math> and the mass of <math>D</math> is <math>289\cdot 240</math>. Hence, the mass of <math>F</math> is <math>289\cdot 49</math>, so the ratio <math>\frac{DE}{EF}=\frac{49}{240}</math>. Extracting gives <math>49+240=\boxed{289}.</math>
  
 
== See also ==
 
== See also ==
Line 26: Line 67:
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 23:37, 11 July 2024

Problem

In $\triangle ABC, AB = 360, BC = 507,$ and $CA = 780.$ Let $M$ be the midpoint of $\overline{CA},$ and let $D$ be the point on $\overline{CA}$ such that $\overline{BD}$ bisects angle $ABC.$ Let $F$ be the point on $\overline{BC}$ such that $\overline{DF} \perp \overline{BD}.$ Suppose that $\overline{DF}$ meets $\overline{BM}$ at $E.$ The ratio $DE: EF$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution 1

In the following, let the name of a point represent the mass located there. Since we are looking for a ratio, we assume that $AB=120$, $BC=169$, and $CA=260$ in order to simplify our computations.

First, reflect point $F$ over angle bisector $BD$ to a point $F'$.

[asy] size(400); pointpen = black; pathpen = black+linewidth(0.7); pair A=(0,0),C=(7.8,0),B=IP(CR(A,3.6),CR(C,5.07)), M=(A+C)/2, Da = bisectorpoint(A,B,C), D=IP(B--B+(Da-B)*10,A--C), F=IP(D--D+10*(B-D)*dir(270),B--C), E=IP(B--M,D--F);pair Fprime=2*D-F; /* scale down by 100x */ D(MP("A",A,NW)--MP("B",B,N)--MP("C",C)--cycle); D(B--D(MP("D",D))--D(MP("F",F,NE))); D(B--D(MP("M",M)));D(A--MP("F'",Fprime,SW)--D); MP("E",E,NE); D(rightanglemark(F,D,B,4)); MP("390",(M+C)/2); MP("390",(M+C)/2); MP("360",(A+B)/2,NW); MP("507",(B+C)/2,NE); [/asy]

As $BD$ is an angle bisector of both triangles $BAC$ and $BF'F$, we know that $F'$ lies on $AB$. We can now balance triangle $BF'C$ at point $D$ using mass points.

By the Angle Bisector Theorem, we can place mass points on $C,D,A$ of $120,\,289,\,169$ respectively. Thus, a mass of $\frac {289}{2}$ belongs at both $F$ and $F'$ because BD is a median of triangle $BF'F$ . Therefore, $CB/FB=\frac{289}{240}$.

Now, we reassign mass points to determine $FE/FD$. This setup involves $\triangle CFD$ and transversal $MEB$. For simplicity, put masses of $240$ and $289$ at $C$ and $F$ respectively. To find the mass we should put at $D$, we compute $CM/MD$. Applying the Angle Bisector Theorem again and using the fact $M$ is a midpoint of $AC$, we find \[\frac {MD}{CM} = \frac {\frac{169}{289}\cdot 260 - 130}{130} = \frac {49}{289}\] At this point we could find the mass at $D$ but it's unnecessary. \[\frac {DE}{EF} = \frac {F}{D} = \frac {F}{C}\cdot\frac {C}{D} = \frac {289}{240}\cdot\frac {49}{289} = \boxed{\frac {49}{240}}\] and the answer is $49 + 240 = \boxed{289}$.

Solution 2

By the Angle Bisector Theorem, we know that $[CBD]=\frac{169}{289}[ABC]$. Therefore, by finding the area of triangle $CBD$, we see that \[\frac{507\cdot BD}{2}\sin\frac{B}{2}=\frac{169}{289}[ABC].\] Solving for $BD$ yields \[BD=\frac{2[ABC]}{3\cdot289\sin\frac{B}{2}}.\] Furthermore, $\cos\frac{B}{2}=\frac{BD}{BF}$, so \[BF=\frac{BD}{\cos\frac{B}{2}}=\frac{2[ABC]}{3\cdot289\sin\frac{B}{2}\cos\frac{B}{2}}.\] Now by the identity $2\sin\frac{B}{2}\cos\frac{B}{2}=\sin B$, we get \[BF=\frac{4[ABC]}{3\cdot289\sin B}.\] But then $[ABC]=\frac{360\cdot 507}{2}\sin B$, so $BF=\frac{240}{289}\cdot 507$. Thus $BF:FC=240:49$.

Now by the Angle Bisector Theorem, $CD=\frac{169}{289}\cdot 780$, and we know that $MC=\frac{1}{2}\cdot 780$ so $DM:MC=\frac{169}{289}-\frac{1}{2}:\frac{1}{2}=49:289$.

We can now use mass points on triangle CBD. Assign a mass of $240\cdot 49$ to point $C$. Then $D$ must have mass $240\cdot 289$ and $B$ must have mass $49\cdot 49$. This gives $F$ a mass of $240\cdot 49+49\cdot 49=289\cdot 49$. Therefore, $DE:EF=\frac{289\cdot 49}{240\cdot 289}=\frac{49}{240}$, giving us an answer of $\boxed{289}.$

Solution 3

Let $\angle{DBM}=\theta$ and $\angle{DBC}=\alpha$. Then because $BM$ is a median we have $360\sin{(\alpha+\theta)}=507\sin{(\alpha-\theta)}$. Now we know \[\sin{(\alpha+\theta)}=\sin{\alpha}\cos{\theta}+\sin{\theta}\cos{\alpha}=\dfrac{DF\cdot BD}{BF\cdot BE}+\dfrac{DE\cdot BD}{BE\cdot BF}=\dfrac{BD(DF+DE)}{BF\cdot BE}\] Expressing the area of $\triangle{BEF}$ in two ways we have \[\dfrac{1}{2}BE\cdot BF\sin{(\alpha-\theta)}=\dfrac{1}{2}EF\cdot BD\] so \[\sin{(\alpha-\theta)}=\dfrac{EF\cdot BD}{BF\cdot BE}\] Plugging this in we have \[\dfrac{360\cdot BD(DF+DE)}{BF\cdot BE}=\dfrac{507\cdot BD\cdot EF}{BF\cdot BE}\] so $\dfrac{DF+DE}{EF}=\dfrac{507}{360}$. But $DF=DE+EF$, so this simplifies to $1+\dfrac{2DE}{EF}=\dfrac{507}{360}=\dfrac{169}{120}$, and thus $\dfrac{DE}{EF}=\dfrac{49}{240}$, and $m+n=\boxed{289}$.

Solution 4 (Overpowered Projective Geometry!!)

Firstly, angle bisector theorem yields $\frac{CD}{AD} = \frac{507}{360} = \frac{169}{120}$. We're given that $AM=MC$. Therefore, the cross ratio

\[(A,C;M,D) = \frac{AM(CD)}{AD(MC)} = \frac{169}{120}\]

We need a fourth point for this cross ratio to be useful, so reflect point $F$ over angle bisector $BD$ to a point $F'$. Then $\triangle BFF'$ is isosceles and $BD$ is an altitude so $DF = DF'$. Therefore,

\[(A,C;M,D) = (F,F';D,E) \implies \frac{FD(EF')}{EF(DF')} = \frac{EF'}{EF} = \frac{169}{120}\]

All that's left is to fiddle around with the ratios:

\[\frac{EF'}{EF} = \frac{ED+DF'}{EF} = \frac{EF+2DE}{EF} = 1\ +\ 2\left(\frac{DE}{EF}\right) \implies \frac{DE}{EF} = \frac{49}{240} \implies \boxed{289}\]

Solution 5 (Menelaus + Mass Points)

Extend $DF$ to intersect with the extension of $AB$ at $G$. Notice that $\triangle{BDF} \cong \triangle{BDG}$, so $GD=DF$. We now use Menelaus on $\triangle{GBF}$, as $A$, $D$, and $C$ are collinear; this gives us $\frac{GA}{BA} \cdot \frac{BC}{FC} \cdot \frac{DF}{GD}=1$. As $GD=DF$, we have $\frac{GA}{AB}=\frac{FC}{BC}$, hence $\frac{GA}{120}=\frac{FC}{169}$. Reflect $G$ over $A$ to $G'$. Note that $\frac{G'A}{BA}=\frac{FC}{BC}$, and reflexivity, hence $\triangle{ABC} \sim \triangle{BG'F}$. It's easily concluded from this that $G'F \parallel AC$, hence $G'F \parallel AD$. As $GD=DF$, we have $AD$ is a midsegment of $\triangle{GG'F}$, thus $G'F = 2AD$. We now focus on the ratio $\frac{BF}{BC}$. From similarity, we have $\frac{BF}{BC}=\frac{G'F}{AC}=\frac{2AD}{AC}$. By the angle bisector theorem, we have $AD:DC=120:169$, hence $AD:AC=120:289$, so $\frac{BF}{BC}=\frac{240}{289}$. We now work out the ratio $\frac{DM}{MC}$. $\frac{DM}{MC}=\frac{CD-MC}{MC}=\frac{CD}{MC}-1=\frac{2CD}{AC}-1=\frac{338}{289}-1=\frac{49}{289}$. We now use mass points on $\triangle{BDC}$. We let the mass of $C$ be $240\cdot 49$, so the mass of $B$ is $49 \cdot 49$ and the mass of $D$ is $289\cdot 240$. Hence, the mass of $F$ is $289\cdot 49$, so the ratio $\frac{DE}{EF}=\frac{49}{240}$. Extracting gives $49+240=\boxed{289}.$

See also

2003 AIME I (Problems • Answer Key • Resources)
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