5
$\begingroup$

enter image description here

Context: I made this question and am contemplating about submitting it as a contest question. The contest is not a large one; its scope does not comprise even our whole class. Its just a friendly, small contest within a small group.
This originally occurred to me when we were asked to prove $\tan \frac A2\tan\frac B2 +\tan \frac B2\tan\frac C2+\tan\frac C2\tan\frac A2=1$ in class. I tried a geometrical proof but failed, but I got the following for my efforts.

Derivation: Given $\triangle ABC$ and its incircle $\Im$ with inradius $r$, we have the known (and easily provable) formula $$\tan \frac A2\tan\frac B2 +\tan \frac B2\tan\frac C2+\tan\frac C2\tan\frac A2=1.\tag{1}\label{1}$$ We know, $IE=IF=IG=r$. Now, $$\tan \frac A2=\frac{r}{AE}=\frac{r}{AF}$$$$\tan \frac B2=\frac{r}{BE}=\frac{r}{BG}$$$$\tan \frac C2=\frac{r}{CG}=\frac{r}{CF}$$ Putting appropriate values in the formula $(1)$, we get $$\frac{r^2}{AE\cdot BE}+ \frac{r^2}{CG\cdot BG}+ \frac{r^2}{AF\cdot CF}=1$$ or $$\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}=\frac{1}{r^2}$$ Finally, using the identities $\Delta=rs, s=\dfrac{a+b+c}{2}, \Delta=\dfrac{abc}{4R}$ (where R is the circumradius) and the sine law we get $$\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}=\dfrac{s^2}{\Delta^2}=\dfrac{(a+b+c)^2}{4\Delta^2}=\dfrac{(a+b+c)^2\cdot 16R^2}{4a^2b^2c^2} $$ Thus, $$\sqrt{\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}}=2R\frac{a+b+c}{abc}$$ $$=2R\frac{2R(\sin A+\sin B+\sin C)}{8R^3\sin A\sin B\sin C}$$$$=\frac{1}{2R}\left(\frac{1}{\sin B\sin C}+ \frac{1}{\sin A\sin C}+ \frac{1}{\sin B\sin A}\right)$$ so that $$2R\sqrt{\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}}=\left(\frac{1}{\sin B\sin C}+ \frac{1}{\sin A\sin C}+ \frac{1}{\sin B\sin A}\right)$$

Question: Prove that for a triangle ABC, $$2R\sqrt{\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}}=\left(\frac{1}{\sin B\sin C}+ \frac{1}{\sin A\sin C}+ \frac{1}{\sin B\sin A}\right)$$

The result looks pretty symmetrical and nice, and also reminds one of Ceva’s theorem because of the segments taken consecutively.

If you were given this question and did not know the above derivation, how would you approach it? Basically, I am looking for alternative solutions to this problem just for reference.

If the question has any flaws, please tell me. Also, I would really like to know how you would rate this question for a high school level informal contest. If any of you can come up with variants of this, especially as inequalities, I would be grateful. Thanks in advance.

$\endgroup$
4
  • $\begingroup$ Have you tried using ceva/trig ceva? $\endgroup$ Commented Jul 14, 2022 at 16:09
  • $\begingroup$ Where are the cevians @TheBestMagician? AIG, BIF and CIE (If that’s what you are referring to) are not straight lines. $\endgroup$ Commented Jul 14, 2022 at 16:12
  • $\begingroup$ My bad, I was being a bit silly. The idea still may work, however, if you draw in the angle bisectors and use ceva that way. You can find the distances between the tangency points and the feet of the angle bisectors. It's probably more useful to use equal tangents, however. Compute $AE$, etc. directly. $\endgroup$ Commented Jul 14, 2022 at 16:15
  • $\begingroup$ >If the question has any flaws, please tell me. Also, I would really like to know how you would rate this question for a high school level informal contest. If any of you can come up with variants of this, especially as inequalities, I would be grateful. Thanks in advance. You can try post the problm on AOPS. The people there are like literally focused on this thing only. $\endgroup$ Commented Jul 22, 2022 at 2:25

2 Answers 2

5
$\begingroup$

Note that $AE=s-a, BE=s-b, BG=s-b, GC=s-c, CF=s-c, FA=s-a$. The expression under the square root becomes $$\sqrt{\frac{1}{(s-a)(s-b)}+\frac{1}{(s-b)(s-c)}+\frac{1}{(s-a)(s-c)}}$$ By Heron's formula, $K=\sqrt{s(s-a)(s-b)(s-c)}\implies \frac{1}{(s-b)(s-c)}=\frac{s(s-a)}{K^2}$, etc. and so the expression under square root becomes $$\frac{1}{K}\sqrt{s(s-a)+s(s-b)+s(s-c)}=\frac{s}{K}$$ So we want to show that $$\frac{2Rs}{K}=\sum_{cyc}\frac{1}{\sin A\sin B}$$ Now, by the extended law of sines, $\frac{a}{\sin A}=2R\implies \frac{1}{\sin A}=\frac{2R}{a}$, so the RHS is $$(2R)^2\sum_{cyc}\frac{1}{ab}=\frac{(2R)^2(a+b+c)}{abc}$$ Thus we want to show $$\frac{s}{K}=\frac{2R(a+b+c)}{abc}$$ Now this is easy because it rearranges to $\frac{1}{K}=\frac{4R}{abc}$, which is $K=\frac{abc}{4R}$ which is true.

$\endgroup$
3
  • $\begingroup$ Nice! Once you reached $\frac sK$ it was basically the result. The rest of my post just consists of algebraic manipulation to enhance beauty. $\endgroup$ Commented Jul 14, 2022 at 16:35
  • $\begingroup$ Yes I left in the algebra because I think it is a little bit cleaner than your derivation $\endgroup$ Commented Jul 14, 2022 at 16:44
  • $\begingroup$ Comment by @albertchan:” Note that AE=s−a,BE=s−b,BG=s−b,GC=s−c,CF=s−c,FA=s−a” For those who don't see it, crucial for the proof. $2s = AB+AC+BC = (AE\!+\!EB)+(AF\!+\!AC)+(BG\!+\!GC)$ $AE=AF\;,\;BE=BG\;,\;CF=CG$. Divide above by 2, we have: $s = AE+BG+CG = AE+a \qquad → AE=s-a$ $s = BE+AF+FC = BE+b \qquad → BE=s-b$ $s = CG+AE+BE = CG+c \qquad → CG=s-c$ $\endgroup$ Commented Jul 15, 2022 at 4:02
3
$\begingroup$

By the result $1$ found in my post, $$\displaystyle x yz=r^{2}(x+y+z),\tag*{(*)} $$

where $r$ is the in-radius of the triangle ABC of sides $a,b$ and $c$ and $a=y+z, b=z+x $ and $c=x+y$.

Dividing the equation (*) by $xyzr^2$ yields the required equation $$ \frac{1}{y z}+\frac{1}{x z}+\frac{1}{x y}=\frac{1}{r^{2}} $$

$\endgroup$
2
  • $\begingroup$ Oh this is quite nice. Congratulations $\endgroup$ Commented Jul 15, 2022 at 3:14
  • $\begingroup$ This is revolutionary. $\endgroup$ Commented Jul 15, 2022 at 7:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .