Let $ABC$ be an acute triangle. If $AD, BE,$ and $CF$ are the altitudes of the triangle $ABC$, prove that $$\text{perimeter of $\triangle{DEF} \leq \text{semiperimeter of $\triangle{ABC.}$}$}$$

We are to prove that the orthic triangle $DEF$'s perimeter is less than or equal to the semi perimeter of $\triangle{ABC}$. Of course, the perimeter of the orthic triangle is minimal among all inscribed triangles in $ABC$, but I don't think that helps us prove the inequality. Also I find it hard to relate the perimeter of the orthic triangle since we don't know very much about its sides.

enter image description here


Okay, from your diagram, let $ H_a=D,H_b=E$, and $H_c=F$. Notice cyclic quadrilateral $AFHE$. Center of a circle circumscribed around $AFHE$ is midpoint of $AH$ with radius of $\rho= \frac{AH}{2}$. It follows that $ EF= 2 \rho \sin \alpha= AH \sin \alpha$ ($\alpha$ is angle at point $A$).

And $AH=\frac{AF}{\sin \beta}= \frac{b \cos \alpha } {\sin \beta}$ and since sine law says

$\frac{a } {\sin \alpha}= \frac{b} {\sin \beta}$

You get that

$EF= a \cos \alpha $

Similary you will find that $DF= b \cos \beta $ etc...

By Chebyshev inequality

$ DE+EF+FD = a \cos \alpha+ b \cos \beta +c \cos \gamma \leq \frac{1}{3} (a+b+c)(\cos \alpha + \cos \beta + \cos \gamma ) \leq \frac{a+b+c}{2} $

Since $\cos \alpha + \cos \beta + \cos \gamma \leq \frac{3}{2} $


If you know Nine-Point circle, it may be a little bit quick:

since $D,E,F$ are on the Nine-point circle, let the radius $R_0, R_1=2R_0,R_1$is the radius of $\triangle ABC$

let $m,n,p$ are the side length of $\triangle DEF, $ the the three angles are $ \pi-2A,\pi-2B,\pi-2C$ $\dfrac{a+b+c}{sinA+sinB+sinC}=2R_1,\dfrac{m+n+p}{sin2A+sin2B+sin2C}=2R_0$

it remains $sin2A+sin2B+sin2C \le sinA+sinB+sinC$

you can use Chebyshev inequality get result as zezanjee's solution or

$sin2A+sin2B=2sinC cos(A-B)\implies sin2A+sin2B+sin2C=cos(B-C)sinA+cos(C-A)sinB+cos(A-B)sinC \le sinA+sinB+sinC$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.