Difference between revisions of "2002 AIME I Problems/Problem 13"

Latest revision as of 22:15, 11 July 2024

Problem

In triangle $ABC$ the medians $\overline{AD}$ and $\overline{CE}$ have lengths $18$ and $27$ , respectively, and $AB=24$ . Extend $\overline{CE}$ to intersect the circumcircle of $ABC$ at $F$ . The area of triangle $AFB$ is $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$ .

Solution 1

$[asy] size(150); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(8); pair A=(0,0), B=(24,0), E=(A+B)/2, C=IP(CR(A,3*70^.5),CR(E,27)), D=(B+C)/2, F=IP(circumcircle(A,B,C),E--C+2*(E-C)); D(D(MP("A",A))--D(MP("B",B))--D(MP("C",C,NW))--cycle); D(circumcircle(A,B,C)); D(MP("F",F)); D(A--D); D(C--F); D(A--F--B); D(MP("E",E,NE)); D(MP("D",D,NE)); MP("12",(A+E)/2,SE,f);MP("12",(B+E)/2,f); MP("27",(C+E)/2,SW,f); MP("18",(A+D)/2,SE,f); [/asy]$

Applying Stewart's Theorem to medians $AD, CE$ , we have:

$\begin{align*} BC^2 + 4 \cdot 18^2 &= 2\left(24^2 + AC^2\right) \\ 24^2 + 4 \cdot 27^2 &= 2\left(AC^2 + BC^2\right) \end{align*}$

Substituting the first equation into the second and simplification yields $24^2 = 2\left(3AC^2 + 2 \cdot 24^2 - 4 \cdot 18^2\right)- 4 \cdot 27^2$ $\Longrightarrow AC = \sqrt{2^5 \cdot 3 + 2 \cdot 3^5 + 2^4 \cdot 3^3 - 2^7 \cdot 3} = 3\sqrt{70}$ .

By the Power of a Point Theorem on $E$ , we get $EF = \frac{12^2}{27} = \frac{16}{3}$ . The Law of Cosines on $\triangle ACE$ gives

$\begin{align*} \cos \angle AEC = \left(\frac{12^2 + 27^2 - 9 \cdot 70}{2 \cdot 12 \cdot 27}\right) = \frac{3}{8} \end{align*}$

Hence $\sin \angle AEC = \sqrt{1 - \cos^2 \angle AEC} = \frac{\sqrt{55}}{8}$ . Because $\triangle AEF, BEF$ have the same height and equal bases, they have the same area, and $[ABF] = 2[AEF] = 2 \cdot \frac 12 \cdot AE \cdot EF \sin \angle AEF = 12 \cdot \frac{16}{3} \cdot \frac{\sqrt{55}}{8} = 8\sqrt{55}$ , and the answer is $8 + 55 = \boxed{063}$ .

Solution 2

Let $AD$ and $CE$ intersect at $P$ . Since medians split one another in a 2:1 ratio, we have

$\begin{align*} AP = 12, PE = 9 \end{align*}$

This gives isosceles $APE$ and thus an easy area calculation. After extending the altitude to $PE$ and using the fact that it is also a median, we find

$\begin{align*} [APE] = \frac{27\sqrt{55}}{4} \end{align*}$

Using Power of a Point, we have

$\begin{align*} EF=\frac{16}{3} \end{align*}$

By Same Height Different Base,

$\begin{align*} \frac{[AFE]}{[APE]}=\frac{[AFE]}{(\frac{27\sqrt{55}}{4})}=\frac{EF}{PE}=\frac{(\frac{16}{3})}{9}=\frac{16}{27} \end{align*}$

Solving gives

$\begin{align*} [AFE] = 4\sqrt{55} \end{align*}$

and

$\begin{align*} [AFB]=2[AFE]=8\sqrt{55} \end{align*}$

Thus, our answer is $8+55=\boxed{063}$ .

Short Solution: Smart Similarity

Use the same diagram as in Solution 1. Call the centroid $P$ . It should be clear that $PE=9$ , and likewise $AP=12$ , $AE=12$ . Then, $\sin \angle AEP = \frac{\sqrt{55}}{8}$ . Power of a Point on $E$ gives $FE=\frac{16}{3}$ , and the area of $AFB$ is $AE * EF* \sin \angle AEP$ , which is twice the area of $AEF$ or $FEB$ (they have the same area because of equal base and height), giving $8\sqrt{55}$ for an answer of $\boxed{063}$ .

Solution 4 (You've Forgotten Power of a Point Exists)

Note that, as above, it is quite easy to get that $\sin \angle AEP = \frac{\sqrt{55}}{8}$ (equate Heron's and $\frac{1}{2}ab\sin C$ to find this). Now note that $\angle FEA = \angle BEC$ because they are vertical angles, $\angle FAE = \angle ECB$ , and $\angle EFA = \angle ABC$ (the latter two are derived from the inscribed angle theorem). Therefore $\Delta AEF$ ~ $\Delta CEB$ and so $FE = \frac{144}{27}$ and $\sin \angle FEA = \frac{\sqrt{55}}{8}$ so the area of $\Delta BFA$ is $8\sqrt{55}$ giving us $\boxed{063}$ as our answer. (One may just get the area via triangle similarity too--this is if you are tired by the end of test and just want to bash some stuff out--it may also serve as a useful check).

~Dhillonr25

Solution 5 (Barycentric Coordinates)

Apply barycentric coordinates on $\triangle ABC$ . We know that $D=\left(0, \frac{1}{2}, \frac{1}{2}\right), E=\left(\frac{1}{2}, \frac{1}{2}, 0\right)$ . We can now get the displacement vectors $\overrightarrow{AD} = \left(1, -\frac{1}{2}, -\frac{1}{2}\right)$ and $\overrightarrow{CE}=\left(-\frac{1}{2}, -\frac{1}{2}, 1\right)$ . Now, applying the distance formula and simplifying gives us the two equations $\begin{align*} 2b^2+2c^2-a^2&=1296 \\ 2a^2+2b^2-c^2&=2916. \\ \end{align*}$ Substituting $c=24$ and solving with algebra now gives $a=6\sqrt{31}, b=3\sqrt{70}$ . Now we can find $F$ . Note that $CE$ can be parameterized as $(1:1:t)$ , so plugging into the circumcircle equation and solving for $t$ gives $t=\frac{-c^2}{a^2+b^2}$ so $F=(a^2+b^2:a^2+b^2:-c^2)$ . Plugging in for $a,b$ gives us $F=(1746:1746:-576)$ . Thus, by the area formula, we have $\frac{[AFB]}{[ABC]}= \left|\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \frac{97}{162} & \frac{97}{162} & -\frac{16}{81} \end{matrix}\right|=\frac{16}{81}.$ By Heron's Formula, we have $[ABC]=\frac{81\sqrt{55}}{2}$ which immediately gives $[AFB]=8\sqrt{55}$ from our ratio, extracting $\boxed{063}$ .

-Taco12

Solution 6 (Law of Cosines + Stewarts)

Since $AD$ is the median, let $BD=BC=x$ . Since $CE$ is a median, $AE=BE=12$ . Applying Power of a Point with respect to point $E$ , we see that $EF=\frac{16}{3}$ . Applying Stewart's Theorem on triangles $\triangle ADC$ and $\triangle ABC$ , we get that $x=3\sqrt{31}$ and $y=3\sqrt{70}$ . The area of $\triangle AFB$ is simply $\frac{1}{2} \cdot \sin{\angle FBA}\cdot FB\cdot AB$ . We know $AB=24$ . Also, we know that $\angle FBA = \angle FCA$ . Then, applying Law of Cosines on triangle $EAC$ , we get that $\cos{\angle FCA}=\frac{3\sqrt{70}}{28}$ which means that $\sin{\angle FCA=\angle FBA}=\frac{\sqrt{154}}{28}$ . Then, applying Stewart's Theorem on triangle $FBC$ with cevian $BE$ allows us to receive that $FB=\frac{4\sqrt{70}}{3}$ . Now, plugging into our earlier area formula, we receive $\frac{1}{2} \cdot \frac{\sqrt{154}}{28} \cdot \frac{4\sqrt{70}}{3} \cdot 24 = 8\sqrt{55}.$ Therefore, the desired answer is $8+55=\boxed{063}$ .

~SirAppel

@@ Line 37: / Line 37: @@
 <cmath>\begin{align*}
 [APE] = \frac{27\sqrt{55}}{4}
-\end{align*}</cmath>
-</center>
-Since <math>P</math> is the centroid, we have that
-<center>
-<cmath>\begin{align*}
-[CBE] = 3 [APE] = \frac{81\sqrt{55}}{4}
 \end{align*}</cmath>
 </center>
@@ Line 54: / Line 48: @@
 <center>
 <cmath>\begin{align*}
-\frac{[EFB]}{[CBE]}=\frac{[EFB]}{(\frac{81\sqrt{55}}{4})}=\frac{(\frac{16}{3})}{27}=\frac{16}{81}
+\frac{[AFE]}{[APE]}=\frac{[AFE]}{(\frac{27\sqrt{55}}{4})}=\frac{EF}{PE}=\frac{(\frac{16}{3})}{9}=\frac{16}{27}
 \end{align*}</cmath>
 </center>
@@ Line 60: / Line 54: @@
 <center>
 <cmath>\begin{align*}
-[EFB] = 4\sqrt{55}
+[AFE] = 4\sqrt{55}
 \end{align*}</cmath>
 </center>
@@ Line 66: / Line 60: @@
 <center>
 <cmath>\begin{align*}
-[AFB]=2[EFB]=8\sqrt{55}
+[AFB]=2[AFE]=8\sqrt{55}
 \end{align*}</cmath>
 </center>
 Thus, our answer is <math>8+55=\boxed{063}</math>.
-'''-Solution by thecmd999'''
+==Short Solution: Smart Similarity==
+Use the same diagram as in Solution 1. Call the centroid <math>P</math>. It should be clear that <math>PE=9</math>, and likewise <math>AP=12</math>, <math>AE=12</math>. Then, <math>\sin \angle AEP = \frac{\sqrt{55}}{8}</math>. Power of a Point on <math>E</math> gives <math>FE=\frac{16}{3}</math>, and the area of <math>AFB</math> is <math>AE * EF* \sin \angle AEP</math>, which is twice the area of <math>AEF</math> or <math>FEB</math> (they have the same area because of equal base and height), giving <math>8\sqrt{55}</math> for an answer of <math>\boxed{063}</math>.
+== Solution 4 (You've Forgotten Power of a Point Exists) ==
+Note that, as above, it is quite easy to get that <math>\sin \angle AEP = \frac{\sqrt{55}}{8}</math> (equate Heron's and <math>\frac{1}{2}ab\sin C</math> to find this). Now note that <math>\angle FEA = \angle BEC</math> because they are vertical angles, <math>\angle FAE = \angle ECB</math>, and <math>\angle EFA = \angle ABC</math> (the latter two are derived from the inscribed angle theorem). Therefore <math>\Delta AEF </math> ~ <math> \Delta CEB</math> and so <math>FE = \frac{144}{27}</math> and <math>\sin \angle FEA = \frac{\sqrt{55}}{8}</math> so the area of <math>\Delta BFA</math> is <math>8\sqrt{55}</math> giving us <math>\boxed{063}</math> as our answer. (One may just get the area via triangle similarity too--this is if you are tired by the end of test and just want to bash some stuff out--it may also serve as a useful check).
+~Dhillonr25
+== Solution 5 (Barycentric Coordinates) ==
+Apply barycentric coordinates on <math>\triangle ABC</math>. We know that <math>D=\left(0, \frac{1}{2}, \frac{1}{2}\right), E=\left(\frac{1}{2}, \frac{1}{2}, 0\right)</math>. We can now get the displacement vectors <math>\overrightarrow{AD} = \left(1, -\frac{1}{2}, -\frac{1}{2}\right)</math> and <math>\overrightarrow{CE}=\left(-\frac{1}{2}, -\frac{1}{2}, 1\right)</math>. Now, applying the distance formula and simplifying gives us the two equations
+<cmath>\begin{align*}
+b^2+2c^2-a^2&=1296 \\
+a^2+2b^2-c^2&=2916. \\
+\end{align*}</cmath>
+Substituting <math>c=24</math> and solving with algebra now gives <math>a=6\sqrt{31}, b=3\sqrt{70}</math>. Now we can find <math>F</math>. Note that <math>CE</math> can be parameterized as <math>(1:1:t)</math>, so plugging into the circumcircle equation and solving for <math>t</math> gives <math>t=\frac{-c^2}{a^2+b^2}</math> so <math>F=(a^2+b^2:a^2+b^2:-c^2)</math>. Plugging in for <math>a,b</math> gives us <math>F=(1746:1746:-576)</math>. Thus, by the area formula, we have<cmath>\frac{[AFB]}{[ABC]}=
+\left|\begin{matrix}
+& 0 & 0 \\
+& 1 & 0 \\
+\frac{97}{162} & \frac{97}{162} & -\frac{16}{81}
+\end{matrix}\right|=\frac{16}{81}.</cmath>By Heron's Formula, we have <math>[ABC]=\frac{81\sqrt{55}}{2}</math> which immediately gives <math>[AFB]=8\sqrt{55}</math> from our ratio, extracting <math>\boxed{063}</math>.
+-Taco12
+==Solution 6 (Law of Cosines + Stewarts)==
+Since <math>AD</math> is the median, let <math>BD=BC=x</math>. Since <math>CE</math> is a median, <math>AE=BE=12</math>. Applying Power of a Point with respect to point <math>E</math>, we see that <math>EF=\frac{16}{3}</math>. Applying Stewart's Theorem on triangles <math>\triangle ADC</math> and <math>\triangle ABC</math>, we get that <math>x=3\sqrt{31}</math> and <math>y=3\sqrt{70}</math>. The area of <math>\triangle AFB</math> is simply <math>\frac{1}{2} \cdot \sin{\angle FBA}\cdot FB\cdot AB</math>. We know <math>AB=24</math>. Also, we know that <math>\angle FBA = \angle FCA</math>. Then, applying Law of Cosines on triangle <math>EAC</math>, we get that <math>\cos{\angle FCA}=\frac{3\sqrt{70}}{28}</math> which means that <math>\sin{\angle FCA=\angle FBA}=\frac{\sqrt{154}}{28}</math>. Then, applying Stewart's Theorem on triangle <math>FBC</math> with cevian <math>BE</math> allows us to receive that <math>FB=\frac{4\sqrt{70}}{3}</math>. Now, plugging into our earlier area formula, we receive <math>\frac{1}{2} \cdot \frac{\sqrt{154}}{28} \cdot \frac{4\sqrt{70}}{3} \cdot 24 = 8\sqrt{55}.</math> Therefore, the desired answer is <math>8+55=\boxed{063}</math>.
+~SirAppel
 == See also ==
@@ Line 77: / Line 99: @@
 [[Category:Intermediate Geometry Problems]]
+{{MAA Notice}}

2002 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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