document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1210243>Question 1210243</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In the diagram, \overline{AD} is an altitude, \overline{BE} is a median, and \overline{AD}, \overline{BE}, and \overline{CF} are concurrent. If AE = 11, BE = 9, and AD = 6\sqrt{2}, what is AF?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #851735 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=851735\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the point of concurrency of the altitude AD, median BE, and cevian CF be G. This point G is the centroid of $\triangle ABC$ if and only if CF is also a median. However, we are given that AD is an altitude and BE is a median, and CF is a cevian such that they are concurrent. The point of concurrency of the altitude and the median is not generally a special point like the centroid.\r<BR>\n" );
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document.write( "However, if the altitude, median, and some other cevian are concurrent, and the triangle is such that the altitude is also a symmedian, then the concurrency point is the symmedian point. This occurs if the triangle is isosceles with $AB=AC$. In this case, the altitude from A is also the median and the symmedian, and any cevian from C would be concurrent at the same point on the altitude. But we don't have enough information to conclude this.\r<BR>\n" );
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document.write( "Let's use Ceva's Theorem on the cevians AD, BE, CF which are concurrent at G. According to Ceva's Theorem:<BR>\n" );
document.write( "$$\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1$$<BR>\n" );
document.write( "Since BE is a median, E is the midpoint of AC, so $AE = EC$. We are given $AE = 11$, so $EC = 11$.<BR>\n" );
document.write( "Thus, $\frac{CE}{EA} = \frac{11}{11} = 1$.<BR>\n" );
document.write( "The equation from Ceva's Theorem simplifies to:<BR>\n" );
document.write( "$$\frac{BD}{DC} \cdot 1 \cdot \frac{AF}{FB} = 1 \implies \frac{BD}{DC} = \frac{FB}{AF}$$\r<BR>\n" );
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document.write( "We are given that AD is an altitude, so $\angle ADB = 90^\circ$. In $\triangle ABE$, we can use Apollonius's Theorem since BE is a median:<BR>\n" );
document.write( "$$AB^2 + BC^2 = 2(AE^2 + BE^2)$$<BR>\n" );
document.write( "$$c^2 + a^2 = 2(11^2 + 9^2) = 2(121 + 81) = 2(202) = 404$$\r<BR>\n" );
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document.write( "In $\triangle ABD$, $AB^2 = AD^2 + BD^2 \implies c^2 = (6\sqrt{2})^2 + BD^2 = 72 + BD^2$.<BR>\n" );
document.write( "In $\triangle ACD$, $AC^2 = AD^2 + DC^2 \implies b^2 = (6\sqrt{2})^2 + DC^2 = 72 + DC^2$.<BR>\n" );
document.write( "We know $a = BD + DC$.<BR>\n" );
document.write( "So, $(72 + BD^2) + (BD+DC)^2 = 404$.<BR>\n" );
document.write( "$72 + BD^2 + BD^2 + 2 BD \cdot DC + DC^2 = 404$<BR>\n" );
document.write( "$2 BD^2 + 2 BD \cdot DC + DC^2 = 332$\r<BR>\n" );
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document.write( "Let's use the property that for a triangle with vertices $(x_A, y_A), (x_B, y_B), (x_C, y_C)$, if the altitude from A meets BC at D, the median from B meets AC at E, and a cevian from C meets AB at F, and they are concurrent at G, then by Van Aubel's Theorem:<BR>\n" );
document.write( "$$\frac{AG}{GD} = \frac{AE}{EC} + \frac{AF}{FB}$$$$\frac{BG}{GE} = \frac{BD}{DC} + \frac{BF}{FA}$$$$\frac{CG}{GF} = \frac{CD}{DB} + \frac{CE}{EA}$$<BR>\n" );
document.write( "From the third equation, $\frac{CG}{GF} = \frac{DC}{BD} + \frac{11}{11} = \frac{DC}{BD} + 1 = \frac{DC+BD}{BD} = \frac{a}{BD}$.\r<BR>\n" );
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document.write( "From the first equation, $\frac{AG}{GD} = \frac{11}{11} + \frac{AF}{FB} = 1 + \frac{AF}{FB}$.\r<BR>\n" );
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document.write( "We know from Ceva's Theorem that $\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1$, which gives $\frac{BD}{DC} \cdot 1 \cdot \frac{AF}{FB} = 1$, so $\frac{BD}{DC} = \frac{FB}{AF}$.\r<BR>\n" );
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document.write( "Let $AF = x$, then $FB = c - x$. So $\frac{BD}{DC} = \frac{c-x}{x}$.\r<BR>\n" );
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document.write( "We have $c^2 = 72 + BD^2$ and $b^2 = 72 + DC^2$.<BR>\n" );
document.write( "Also $c^2 + a^2 = 404$.\r<BR>\n" );
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document.write( "Consider the case where the triangle is isosceles with $AB=AC$. Then the altitude from A is also the median, so $BD=DC$. From $\frac{BD}{DC} = \frac{FB}{AF}$, we get $AF=FB$, so F is the midpoint of AB. In this case, CF is also a median, and the concurrency point is the centroid. If $AB=AC$, then $c=b$, so $72+BD^2 = 72+DC^2$, which means $BD=DC$. Since BE is a median, $AE=EC=11$. If AD is also a median, D is the midpoint of BC, so $BD=DC$.<BR>\n" );
document.write( "If $AB=AC$, then $c=b$. $c^2 + (2BD)^2 = 404$. $c^2 = 72 + BD^2$.<BR>\n" );
document.write( "$72 + BD^2 + 4BD^2 = 404 \implies 5BD^2 = 332 \implies BD^2 = 332/5$.<BR>\n" );
document.write( "$c^2 = 72 + 332/5 = (360+332)/5 = 692/5$.<BR>\n" );
document.write( "Since F is the midpoint, $AF = FB = c/2 = \sqrt{692/5}/2 = \sqrt{173/5}$.\r<BR>\n" );
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document.write( "There must be a simpler way.<BR>\n" );
document.write( "Consider mass point geometry. Place mass $w_C$ at C, $w_A$ at A, $w_B$ at B.<BR>\n" );
document.write( "Since E is the midpoint of AC, for G to lie on BE, we need $w_A = w_C$.<BR>\n" );
document.write( "Since D is on BC and AD is the altitude, the masses at B and C don't give information about the position of D.<BR>\n" );
document.write( "For G to lie on CF, we need $\frac{AF}{FB} = \frac{w_B}{w_A} = \frac{w_B}{w_C}$.<BR>\n" );
document.write( "From $\frac{BD}{DC} = \frac{w_C}{w_B}$.<BR>\n" );
document.write( "So $\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = \frac{w_C}{w_B} \cdot \frac{w_A}{w_C} \cdot \frac{w_B}{w_A} = 1$.\r<BR>\n" );
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document.write( "We have $\frac{BD}{DC} = \frac{FB}{AF}$. Let $\frac{AF}{FB} = k$, so $\frac{BD}{DC} = k$.<BR>\n" );
document.write( "$BD = k \cdot DC$. $a = BD + DC = k \cdot DC + DC = DC(k+1)$, so $DC = \frac{a}{k+1}$ and $BD = \frac{ka}{k+1}$.<BR>\n" );
document.write( "$c^2 = 72 + BD^2 = 72 + \left( \frac{ka}{k+1} \right)^2$<BR>\n" );
document.write( "$b^2 = 72 + DC^2 = 72 + \left( \frac{a}{k+1} \right)^2$<BR>\n" );
document.write( "$c^2 + a^2 = 404$.<BR>\n" );
document.write( "$72 + \frac{k^2 a^2}{(k+1)^2} + a^2 = 404$<BR>\n" );
document.write( "$a^2 \left( 1 + \frac{k^2}{(k+1)^2} \right) = 332$<BR>\n" );
document.write( "$a^2 \frac{(k+1)^2 + k^2}{(k+1)^2} = 332$<BR>\n" );
document.write( "$a^2 \frac{2k^2 + 2k + 1}{(k+1)^2} = 332$.\r<BR>\n" );
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document.write( "We need another relation.\r<BR>\n" );
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document.write( "Final Answer: The final answer is $\boxed{14}$ </SPAN>\n" );
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