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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. | ||
− | + | 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>. | |
− | + | 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 | |
− | |||
<cmath> | <cmath> | ||
− | \frac {MD}{CM} = \frac { | + | \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> | 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> | Solving for <math>BD</math> yields <cmath>BD=\frac{2[ABC]}{3\cdot289\sin\frac{B}{2}}.</cmath> | ||
Line 34: | Line 36: | ||
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> | 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+2DF}{EF} = 1\ +\ 2\left(\frac{DF}{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 == |
Revision as of 16:40, 28 June 2024
Contents
Problem
In and
Let
be the midpoint of
and let
be the point on
such that
bisects angle
Let
be the point on
such that
Suppose that
meets
at
The ratio
can be written in the form
where
and
are relatively prime positive integers. Find
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 ,
, and
in order to simplify our computations.
First, reflect point over angle bisector
to a point
.
![[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]](http://latex.artofproblemsolving.com/d/5/2/d52650d1c43c62a6df30fc3e7b09651eb4cf6b37.png)
As is an angle bisector of both triangles
and
, we know that
lies on
. We can now balance triangle
at point
using mass points.
By the Angle Bisector Theorem, we can place mass points on of
respectively. Thus, a mass of
belongs at both
and
because BD is a median of triangle
. Therefore,
.
Now, we reassign mass points to determine . This setup involves
and transversal
. For simplicity, put masses of
and
at
and
respectively. To find the mass we should put at
, we compute
. Applying the Angle Bisector Theorem again and using the fact
is a midpoint of
, we find
At this point we could find the mass at
but it's unnecessary.
and the answer is
.
Solution 2
By the Angle Bisector Theorem, we know that . Therefore, by finding the area of triangle
, we see that
Solving for
yields
Furthermore,
, so
Now by the identity
, we get
But then
, so
. Thus
.
Now by the Angle Bisector Theorem, , and we know that
so
.
We can now use mass points on triangle CBD. Assign a mass of to point
. Then
must have mass
and
must have mass
. This gives
a mass of
. Therefore,
, giving us an answer of
Solution 3
Let and
. Then because
is a median we have
. Now we know
Expressing the area of
in two ways we have
so
Plugging this in we have
so
. But
, so this simplifies to
, and thus
, and
.
Solution 4 (Overpowered Projective Geometry!!)
Firstly, angle bisector theorem yields . We're given that
. Therefore, the cross ratio
We need a fourth point for this cross ratio to be useful, so reflect point over angle bisector
to a point
. Then
is isosceles and
is an altitude so
. Therefore,
All that's left is to fiddle around with the ratios:
Solution 5 (Menelaus + Mass Points)
Extend to intersect with the extension of
at
. Notice that
, so
. We now use Menelaus on
, as
,
, and
are collinear; this gives us
. As
, we have
, hence
. Reflect
over
to
. Note that
, and reflexivity, hence
. It's easily concluded from this that
, hence
. As
, we have
is a midsegment of
, thus
. We now focus on the ratio
. From similarity, we have
. By the angle bisector theorem, we have
, hence
, so
. We now work out the ratio
.
. We now use mass points on
. We let the mass of
be
, so the mass of
is
and the mass of
is
. Hence, the mass of
is
, so the ratio
. Extracting gives
See also
2003 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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