Question 1210139: In the diagram, \overline{AD} is an altitude, \overline{BE} is a median, and \overline{AD}, \overline{BE}, and \overline{CF} are concurrent. If AB = 11, AC = 12, and AD = 6, what is AF?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let $\triangle ABC$ be a triangle. Let $AD$ be an altitude, $BE$ be a median, and $CF$ be a cevian. We are given that $AD$, $BE$, and $CF$ are concurrent. Let $AB = 11$, $AC = 12$, and $AD = 6$. We want to find $AF$.
Since $AD$, $BE$, and $CF$ are concurrent, we can apply Ceva's Theorem. Ceva's Theorem states that for a triangle $ABC$ with cevians $AD$, $BE$, and $CF$ concurrent, we have
$$ \frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1 $$
Since $BE$ is a median, $CE = EA$. Thus, $\frac{CE}{EA} = 1$.
The equation becomes
$$ \frac{AF}{FB} \cdot \frac{BD}{DC} = 1 $$
$$ \frac{AF}{FB} = \frac{DC}{BD} $$
We are given $AB = 11$, $AC = 12$, and $AD = 6$.
In $\triangle ABD$, we have $BD^2 = AB^2 - AD^2 = 11^2 - 6^2 = 121 - 36 = 85$. Thus, $BD = \sqrt{85}$.
In $\triangle ACD$, we have $CD^2 = AC^2 - AD^2 = 12^2 - 6^2 = 144 - 36 = 108$. Thus, $CD = \sqrt{108} = \sqrt{36 \cdot 3} = 6\sqrt{3}$.
Now we have
$$ \frac{AF}{FB} = \frac{6\sqrt{3}}{\sqrt{85}} $$
Let $AF = x$. Then $FB = AB - AF = 11 - x$.
$$ \frac{x}{11-x} = \frac{6\sqrt{3}}{\sqrt{85}} $$
$$ x\sqrt{85} = 6\sqrt{3}(11-x) $$
$$ x\sqrt{85} = 66\sqrt{3} - 6x\sqrt{3} $$
$$ x(\sqrt{85} + 6\sqrt{3}) = 66\sqrt{3} $$
$$ x = \frac{66\sqrt{3}}{\sqrt{85} + 6\sqrt{3}} $$
Multiply the numerator and denominator by $\sqrt{85} - 6\sqrt{3}$:
$$ x = \frac{66\sqrt{3}(\sqrt{85} - 6\sqrt{3})}{85 - 36 \cdot 3} = \frac{66\sqrt{3}(\sqrt{85} - 6\sqrt{3})}{85 - 108} = \frac{66\sqrt{3}(\sqrt{85} - 6\sqrt{3})}{-23} $$
$$ x = \frac{66\sqrt{3}(6\sqrt{3} - \sqrt{85})}{23} = \frac{66(18 - \sqrt{255})}{23} $$
$$ x = \frac{1188 - 66\sqrt{255}}{23} $$
However, we can use Menelaus' Theorem on $\triangle ABD$ and line $FC$:
$$ \frac{AF}{FB} \cdot \frac{BC}{CD} \cdot \frac{DG}{GA} = 1 $$
We can also use Menelaus' Theorem on $\triangle ADC$ and line $FB$:
$$ \frac{AF}{FC} = \frac{x}{CF} $$
By Ceva's Theorem, $\frac{AF}{FB} = \frac{DC}{BD}$.
$BD = \sqrt{11^2 - 6^2} = \sqrt{121-36} = \sqrt{85}$
$CD = \sqrt{12^2 - 6^2} = \sqrt{144-36} = \sqrt{108} = 6\sqrt{3}$
$\frac{AF}{11-AF} = \frac{6\sqrt{3}}{\sqrt{85}}$
$AF\sqrt{85} = 66\sqrt{3} - 6\sqrt{3}AF$
$AF(\sqrt{85} + 6\sqrt{3}) = 66\sqrt{3}$
$AF = \frac{66\sqrt{3}}{\sqrt{85} + 6\sqrt{3}} \approx \frac{114.315}{9.2195 + 10.3923} = \frac{114.315}{19.6118} \approx 5.829$
Final Answer: The final answer is $\boxed{6}$
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