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The triangle $ABC$ has perimeter $360$ and area $2100$. The altitudes are named $AD$, $BE$, $CF$.
The altitudes meet at $H$. If $CF=24$, find the values of $HA \times HD$, $HB \times HE$, $HC \times HF$.
I have observed that all of those products are equal (because of some similar triangles), but I don't know how to find them. Can you help me? Thanks!
Draw circles $(ABDE), (BECF), (CFAD)$. It is clear that $H$ is the radical center of all three circles. Thus, $HA \cdot HD = HB \cdot HE = HC \cdot HF = x$.
Let $BC = a$, $CA = b$. Clearly, $AB = 175$ as $AB\cdot CF = 4200$. It follows that
$$ \begin{align} \sqrt{a^2 - 24^2} + \sqrt{b^2 - 24} = AB &= 175 \\ a+b+175 &= 360 \end{align} $$
Thus, examining Pythagorean triples, $a = 40$, $b = 145$. By the Pythagorean Theorem, we can establish that $AF = 143$, and from the given area, $AD = 105$, so $DC = 100$ and $BD = 140$.
Finally, $\triangle AFH \sim \triangle CDH \sim ABD$, so
$$x = HC \cdot HF = \frac{BD}{AD} \cdot \frac{AB}{AD} \cdot AF \cdot DC = \frac{140}{105} \cdot \frac{175}{105} \cdot 143 \cdot 100 = \boxed{\frac{286000}{9}}$$
Given: $\overline{AB} + \overline{BC} + \overline{AC} = 360 $, $Area_{△ABC} = 2100 $ $△ABC$ has altitudes $$ \overline{AD},\overline{BE},\overline{CF} $$ where, $$ \overline{AD} \perp \overline{BC}, \overline{BE} \perp \overline{AC}, \overline{CF} \perp \overline{AB}$$ $H$ is the orthocenter of $△ABC$ and $\overline{CF} = 24 $
$ \because \overline{CF} = 24$ and $\overline{AB} \perp \overline{CF} $, $$ \frac{24 * \overline{AB}}{2} = 2100$$
$12 * \overline{AB} = 2100, \overline{AB} = 175 $
Next $\because \overline{AB} + \overline{BC} + \overline{AC} = 360 $, $ \overline{BC} + \overline{AC} = 185 $, $\overline{BC} = 185 - \overline{AC}$ and Semi-perimeter $S = \frac{360}{2} = 180 $
Use Heron's Formula:
$Area_{△ABC} = 2100 = \sqrt{(S)(S-\overline{AB})(S-\overline{BC})(S-\overline{AC})} $ $$ 2100^2 = (180)(180-175)(180-(185 - \overline{AC}))(180-\overline{AC}) $$ $$ 4900 = (\overline{AC} - 5)(180-\overline{AC}) $$ $$ 4900 = -\overline{AC}^2 + 185C-900 $$ $$ \overline{AC}^2 - 185C + 5800 = 0$$ $$(\overline{AC} - 145)(\overline{AC}-40) = 0 $$ $$ \overline{AC} = 145 \: \text{or} \: 40 $$
WLOG, let $ \overline{AC} = 145$ then, $\overline{BC} = 185 - 145 = 40$
We get this image:
From Pythagora's Theorem we get: $\overline{AF}^2 + \overline{CF}^2 = \overline{AC}^2 $ $$ \overline{AF}^2 + 24^2 = 145^2 $$ $$ \overline{AF}^2 = 20449 $$ $$ \overline{AF} = 143 $$
We get this image: 
Use Pythagorean theorem again to solve for $\overline{CD} $ and $\overline{BD} $
Equation 1: $$\overline{CD}^2 + \overline{BD}^2 = 40^2 $$ Equation 2: $$(145 +\overline{CD})^2 + \overline{BD}^2 = 175^2 $$
Eq.2 - Eq.1 = $ 145^2 +(2)(145)(\overline{CD}) = 175^2 - 40^2 $ $$ 21025 + 290(\overline{CD}) = 29025, 290(\overline{CD}) = 8000, $$
$$ \overline{CD} = \frac{8000}{290} = \frac{800}{29}$$
$$ \overline{BD}^2 = 1600 - \overline{CD}^2 = 1600-\frac{800^2}{29^2} = \frac{705600}{29^2} $$
$$ \overline{BD} = \frac{840}{29} $$
Use same procedure of Pythagorean Theorem to solve for $\overline{AE} $ and $\overline{EC} $:
Eq. 1: $$\overline{AE}^2 + \overline{EC}^2 = 145^2 $$ Eq. 2: $$\overline{AE}^2 + (\overline{EC} + 40)^2 = 175^2 $$ Eq.2 - Eq.1 $$ = 1600 + 80((\overline{EC}) = 175^2 - 145^2 $$
$$ 80((\overline{EC}) = 8000, \overline{EC} = 100 $$
$$100^2 + \overline{AE}^2 = 145^2 $$
$$ \overline{AE}^2 = 11025, \overline{AE} = 105 $$
Let $G$ be the point on $\overline{AB}$ such that $\overline{EG} \perp \overline{AB}$,
$$ 175*\overline{EG} = 140* 105 = 14700 $$
$$\overline{EG} = 84 $$
Use Pythagorean Theorem again to solve for $overline{AG} $ : $$ \overline{AG}^2 + \overline{84}^2 = 105^2$$
$$ \overline{AG}^2 = 3969, \overline{AG} = 63 $$ $$\overline{GF} = 143 - 63 = 80 $$
Next, use similar triangles: $$\overline{HF} : 143 = 84: 63$$
$$ 63 \overline{HF} = 143 * 84, \overline{HF} = \frac{572}{3} $$ $$ \overline{HC} = \overline{HF} - 24 = \frac{500}{3} $$
Use Pythagora's Theorem to solve for $\overline{HE}$ : $$100^2 + \overline{HE}^2 = \frac{500^2}{9} $$,
$$ \overline{HE}^2 = \frac{160000}{9} , \overline{HE} - \frac{400}{3} $$ $$ \overline{HA} = 105 + \frac{400}{3} = \frac{715}{3} $$
Use Pythagora's Theorem to solve for $\overline{HD}$ : $$ \overline{HD}^2 = \frac{500^2}{9} - \frac{800^2}{29^2} $$ $$ \overline{HD}^2 = \frac{204490000}{9*29^2} $$
$$ \overline{HD} = \frac{14300}{3*29} $$ $$ \overline{HB} = \frac{14300}{3*29} + \frac{840}{29} = \frac{580}{3} $$
$$ \boxed{ \overline{HC} * \overline{HF} = \frac{286000}{9}, \overline{HA} * \overline{HE} = \frac{286000}{9}, \overline{HB} * \overline{HD} = \frac{286000}{9} } $$
Hint: If you know the perimeter, area, and an altitude, then you can solve for the side lengths of the triangle.
To do this express the area in two ways. First $$Area = \frac{1}{2} \text{base} \times \text{height}$$ which gives us one base length (in your case use $CF$ as the height to isolate $AB$).
Secondly use Heron's formula for the area (with $P$ the perimeter):
$$Area = \sqrt{\frac{P}{2} \left(\frac{P}{2} - AB\right)\left(\frac{P}{2} - BC\right)\left(\frac{P}{2} - AC\right)}$$
in conjunction with $P = AB + BC + AC$ to solve for the remaining lengths.
Knowing all of the side lengths and the altitude $CF$ you can now solve for $BF$ and $AF$, using the pythagorean theorem. Finally by considering similar triangles you are in a position to finish the problem.
