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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!

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  • $\begingroup$ What does, "The altitudes of A, B, C found BC, CA, AB at D, E, F" mean? Are you saying that the altitude to BC is AD, the altitude to AC is BE, and the altitude to AD is CF? $\endgroup$
    – saulspatz
    Jun 3, 2018 at 18:44
  • $\begingroup$ @saulspatz, yes. $\endgroup$ Jun 3, 2018 at 18:47
  • $\begingroup$ @saulspatz do you mean altitude to AB is CF ??? $\endgroup$
    – ray lin
    Jun 3, 2018 at 20:29
  • $\begingroup$ @raylin Yes, I did. $\endgroup$
    – saulspatz
    Jun 3, 2018 at 21:57
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    $\begingroup$ If $CF$ is an altitude then Area = $\frac 12 AB*CF$ so $\frac 12 AB*24 = 2100$ and $AB = 175$. You can use trig (law of sins? cos?) to figure out what $AC + CB = 360 - 175 = 185$ each are. $\endgroup$
    – fleablood
    Jun 3, 2018 at 22:07

3 Answers 3

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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

$$x = HC \cdot HF = \frac{BD}{AD} \cdot \frac{AB}{AD} \cdot AF \cdot DC = \frac{140}{105} \cdot \frac{175}{105} \cdot 143 \cdot 100 = \boxed{\frac{286000}{9}}$$

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  • $\begingroup$ Comforting that the two answers agree. $\endgroup$ Jun 4, 2018 at 3:52
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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:

Image of Triangle

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: Pythagora's Theorem

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 $\overline{EG} \perp \overline{AB}$,

$$ 175*\overline{EG} = 140* 105 = 14700 $$

$$\overline{EG} = 84 $$

Progress

Use Pythagorean Theorem again to solve for $overline{AG} $ : $$ \overline{AG}^2 + \overline{84}^2 = 105^2$$

$$ \overline{AG}^2 = 3969, \overline{AG} = 63 $$ $$\overline{GF} = 143 - 63 = 80 $$

more Progress

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 $\overline{HD}$ : $$ \overline{HD}^2 = \frac{500^2}{9} - \frac{800^2}{29^2} $$ $$ \overline{HD}^2 = \frac{204490000}{9*29^2} $$

$$ \overline{HD} = \frac{14300}{3*29} $$ $$ \overline{HB} = \frac{14300}{3*29} + \frac{840}{29} = \frac{580}{3} $$

$$ \boxed{ \overline{HC} * \overline{HF} = \frac{286000}{9}, \overline{HA} * \overline{HE} = \frac{286000}{9}, \overline{HB} * \overline{HD} = \frac{286000}{9} } $$

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  • $\begingroup$ Comforting that the two answers agree. $\endgroup$ Jun 4, 2018 at 3:53
  • $\begingroup$ Thanks for your hard work! :) $\endgroup$ Jun 5, 2018 at 15:20
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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.

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