Proving Altitudes of Triangle can never form a Triangle

Question

Prove that Altitudes of Triangle can never form a Triangle

My try: we have altitudes proportional to reciprocals of sides of given triangle

Let $a,b,c$ are sides we have

$$a+b \gt c $$

$$b+c \gt a$$ and $$c+a \gt b$$

Now if $\frac{1}{a}$, $\frac{1}{b}$ and $\frac{1}{c}$ are sides of another Triangle we need to show that

$\frac{1}{a}+\frac{1}{b}$ Can never be Greater than $\frac{1}{c}$

Any hint?

Answer 1

Let $a \geq b \geq c$ be the sides of a triangle. If $p,q,r$ are the altitudes of the triangle incident at $a,b,c$ respectively, then $$ \frac 12 ap = \frac 12 bq = \frac 12 cr = \Delta, $$ where $\Delta$ is the area of the triangle formed by $a,b,c$. As a consequence, $$ p:q:r = \frac{1}{a} : \frac 1b : \frac 1c. $$ Note that, by scaling a triangle if necessary, we see that $p,q,r$ form a triangle if and only if $\frac 1a,\frac 1b$ and $\frac 1c$ do.

Observe that $\frac 1a \leq \frac 1b \leq \frac 1c$. Therefore, the three quantities can be the sides of a triangle if and only if $\frac 1c < \frac 1b + \frac 1a$, which happens if and only if $c(a+b) > ab$.

Which means that there may be triangles for which the altitudes do actually form a triangle, for example if $a=b=c$, and there are examples where triangles are not formed e.g. if $c(a+b) \leq ab$. You can check that $3,3,1$ is a triangle, whose altitudes would not form a triangle since $1(3+3) = 6$ while $3 \times 3= 9$.

Answer 2

The altitudes of $\triangle ABC$ can form a triangle if and only if

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}>2\max\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$$

where $a,b,c$ are the sides of $\triangle ABC.$

Answer 3

The condition for a triangle is $c(a+b) > ab$.

For the 3-4-5 triangle, this is $5(3+4) > 3\cdot 4$ which is true.

So the altitudes do form a triangle in this case.

Note that you can write the condition as

$\begin{array}\\ 0 &\gt ab-c(a+b)\\ &= ab-c(a+b)+c^2-c^2\\ &= (a-c)(b-c)-c^2\\ \text{or}\\ c^2 &\gt (a-c)(b-c)\\ &= (c-a)(c-b)\\ \end{array} $

This is true if $c$ is the largest side of the triangle, so it is always true.

Therefore, the altitudes can always form a triangle.

Answer 4

The altitudes can form a triangle for some but not all triangles $ABC$.

I. In isosceles triangle $ABC$, with altitudes $AD$, $BE$, $CF$, let $AC=2BC$.

Since$$\triangle ADC\sim\triangle BEC$$then$$\frac{AD}{BE}=\frac{AC}{BC}$$hence$$AD=2BE$$And since $BE=CF$, then $$AD=BE+CF$$and altitudes $AD$, $BE$, $CF$ cannot form a triangle when $\cos \angle ACB=\frac{DC}{AC}=.25$, that is for$$\angle ACB\ge\arccos .25=75.52^o$$But if $AC<2BC$, that is if $\angle ACB<75.52^O$, then since$$AD<2BE=BE+CF$$the three altitudes can form a triangle.

II. The altitudes of a right triangle can form a triangle only if the least angle is not much less than $30^o$.

In triangle $ABC$ with right angle at $C$, suppose $BC>AC$, and hence the least angle is at $B$. Points $E$, $D$ are coincident with $C$. Thus the altitudes can form a triangle only if$$AD+CF>BE$$

Constructing $FG\parallel CB$ and $\angle CBG=\angle FCB$, draw circle with radius $BG=CF$ making $BH=CF$. And let $AD$ be such that $AD=DH$. Then$$AD+CF=BE$$and the three altitudes cannot form a triangle. With $J$ the midpoint of $AB$, evidently $$AD<\frac12AB$$and hence$$\angle ABC<30^o$$

Then keeping $BD$ fixed it is clear that if we decrease $AD$, thereby also decreasing $CF$, then$$AD+CF<BE$$ and again the altitudes cannot form a triangle. But if instead we increase $AD$, hence also increasing $CF$, then$$AD+CF>BE$$and the altitudes can form a triangle. And they will continue to form a triangle until $AD$ has to $BE$ the ratio which $BE$ formerly had to $AD$, i.e. until the angles at $A$ and $B$ are the reverse of what they are in the figure. But if $AD$ is made greater still, the altitudes will no longer form a triangle.

Thus the altitudes of a right triangle can form a triangle if the complementary acute angles are between slightly greater than $60^o$, and slightly less than $30^o$.

It seems clear from just these two types of cases, that the altitudes of a triangle can form a triangle neither always nor never, but only under certain conditions.