Problem 4020 from Crux Mathematicorum

Statement

Problem 4020 from Crux Mathematicorum

Solution 1

Denote by $\alpha,\;$ $\beta,\;$ $\gamma\;$ the angles $BAC,\;$ $ABC,\;$ and $ACB,\;$ respectively; and let $r\;$ be the inradius of $\Delta ABC.

From the quadrilateral $PIMB,\;$ $\angle PIM = 180^{\circ}-\beta,\;$ whence

$\displaystyle [PIM]=\frac{PI\cdot MI}{2}\sin\angle PIM=\frac{r^2}{2}\sin (180^{\circ}-\beta)=\frac{r^2}{2}\sin \beta.$

Similarly, we calculate $[MIN]\;$ and $[NIP],\;$ and get

$\displaystyle\begin{align} [MNP]&=[PIM]+[MIN]+[NIP]\\ &=\frac{r^2}{2}\left(\sin\alpha +\sin\beta +\sin\gamma\right). \end{align}\;\;\;\;\;\text{(1)}$

On the other hand, we have $\displaystyle\angle FID=\angle AIC=180^{\circ}-\frac{\alpha}{2}-\frac{\gamma}{2},\;$ and so

$\displaystyle [FID]=\frac{ID\cdot IF}{2}\sin\angle FID=\frac{ID\cdot IF}{2}\sin\angle\frac{\alpha+\gamma}{2}.$

Similarly, calculate $[\Delta EIF]\;$ and $[\Delta DIE].\;$ We have

$\displaystyle\begin{align} [DEF]&=[DIE]+[EIF]+[FID]\\ &=\frac{ID\cdot IE}{2}\sin\frac{\alpha+\beta}{2}+\frac{IE\cdot IF}{2}\sin\frac{\beta+\gamma}{2}+\frac{ID\cdot IF}{2}\sin\frac{\alpha+\gamma}{2}. \end{align}\;\;\;\;\;\text{(2)}$

Triangles $PIF,\;$ $MID,\;$ and $NIF\;$ are all right-angled, implying that $IF\ge r,\;$ $ID\ge r,\;$ $IE\ge r.\;$ Hence, from (2),

$\displaystyle [\Delta DEF]\ge \frac{r^2}{2}\left(\sin\frac{\alpha+\beta}{2}+\sin\frac{\beta+\gamma}{2}+\sin\frac{\alpha+\gamma}{2}\right).$

Comparing this to (1), in order to have $[MNP]\le [DEF],\;$ suffice it to show that

$\displaystyle \sin\alpha +\sin\beta +\sin\gamma\le\sin\frac{\alpha+\beta}{2}+\sin\frac{\beta+\gamma}{2}+\sin\frac{\alpha+\gamma}{2}.\;\;\;\;\;(3)$

However, we know that

$\displaystyle \sin\alpha +\sin\beta=2\sin\frac{\alpha +\beta}{2}\cos\frac{\alpha-\beta}{2}\le 2\sin\frac{\alpha +\beta}{2}.$

Similarly, $\displaystyle \sin\beta +\sin\gamma\le 2\sin\frac{\beta +\gamma}{2}\;$ and $\displaystyle \sin\alpha +\sin\gamma\le 2\sin\frac{\alpha +\gamma}{2}.\;$ Adding up proves (3) and, therefore, the required inequality.

The equality occurs for $\alpha=\beta=\gamma.$

Solution 2

With $R\;$ the circumradius, $O\;$ the circumcenter, $s\;$ the semiperimeter, we have the following sequence:

$\displaystyle [\Delta MNP]=\frac{R^2-OI^2}{4R^2}\cdot [ABC]=\frac{2Rr}{4R^2}=\frac{r}{4R^2}\cdot [\Delta ABC]\\ \displaystyle [\Delta DEF]=\frac{2abc}{(a+b)(b+c)(c+a)}\cdot [ABC]\\ \displaystyle \frac{r}{2R}\le \frac{2abc}{(a+b)(b+c)(c+a)}\\ \displaystyle\begin{align} \frac{r}{2R}&=\frac{[\Delta ABC]}{s}:\frac{2abc}{4[\Delta ABC]}=\frac{2[\Delta ABC]^2}{sabc}\\ &=\frac{2s(s-a)(s-b)(s-c)}{sabc}\\ &=\frac{2(s-a)(s-b)(s-c)}{abc}. \end{align}$

Thus suffice it to prove that

$\displaystyle\frac{2(s-a)(s-b)(s-c)}{abc}\le\frac{2abc}{(a+b)(b+c)(c+a)},$

which is equivalent to

$(s-a)(s-b)(s-c)(a+b)(b+c)(c+a)\le (abc)^2.$

Let $s-a=x\gt 0,\;$ $s-b=y\gt 0,\;$ $s-c=z\gt 0,\;$ then $a=y+z,\;$ $b=x+z,\;$ $c=x+y,\;$ and the above inequality becomes

$xyz(x+y+2z)(x+2y+z)(2x+y+z)\le (x+y)^2(y+z)^2(z+x)^2.$

By the AM-GM inequality,

$\displaystyle\begin{align}(x+y+2z)(x+2y+z)&(2x+y+z)\\ &\le \left(\frac{(x+y+2z)+(x+2y+z)+(2x+y+z)}{3}\,\right)^3\\ &=\left(\frac{4(x+y+z)}{3}\right)^3\\ \end{align}$

Thus, if we can prove that

$\displaystyle xyz\left(\frac{4(x+y+z)}{3}\right)^3\le (x+y)^2(y+z)^2(z+x)^2$

the problem will be solved. The inequality is equivalent to

$\displaystyle \ln x+\ln y+\ln z+3\ln\frac{x+y+z}{3}\le 2\ln\frac{x+y}{2}+2\ln\frac{y+z}{2}+2\ln\frac{z+x}{2}.$

This is Tiberiu Popoviciu's inequality

$\displaystyle f(x)+f(y)+f(z)+3f\left(\frac{x+y+z}{3}\right)\ge 2f\left(\frac{x+y}{2}\right)+2f\left(\frac{y+z}{2}\right)+2f\left(\frac{z+x}{2}\right)$

for the convex function $f(x)=-\ln (x)\;$ which proves the result.

Solution 3

We'll use the barycentric coordinates. The area of a triangle $T\;$ whose vertices have homogeneous barycentric coordinates $(x_1:y_1:z_1),\;$ $(x_2:y_2:z_2),\;$ $(x_3:y_3:z_3),\;$ (relative to $\Delta ABC)\;$ is given by

$\displaystyle [T]=\frac{[\Delta ABC]}{(x_1+y_1+z_1)(x_2+y_2+z_2)(x_3+y_3+z_3)}\left|\begin{array} \,x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{array}\right|.$

For a triangle formed by the feet of three cevians that concur at point $(x:y:z)\;$ that formula becomes

$\displaystyle [T]=\frac{[\Delta ABC]}{(y+z)(z+x)(x+y)}\left|\begin{array} \,0&y&z\\x&0&z\\x&y&0\end{array}\right|=\frac{2xyz[\Delta ABC]}{(y+z)(z+x)(x+y)}.$

In particular, since $I=(a:b:c),\;$

$\displaystyle [\Delta DEF]=\frac{2abc[\Delta ABC]}{(a+b)(b+c)(c+a)}.$

The vertices of the intouch triangle $MNP\;$ have barycentric coordinates $(0:s-b:s-c),\;$ $(s-a:0:s-c),\;$ $(s-a:s-b:0),\;$ the corresponding cevians meet at the Gergonne point, and, therefore,

$\displaystyle [\Delta MNP]=\frac{2(s-a)(s-b)(s-c)[\Delta ABC]}{abc}$

which reduces the problem to proving

$\displaystyle \frac{(s-a)(s-b)(s-c)}{abc}\le\frac{abc}{(a+b)(b+c)(c+a)}.$

This, in turn, is equivalent to

$\displaystyle \frac{(a+b)^2(s-a)(s-b)}{abc^2}\cdot\frac{(b+c)^2(s-b)(s-c)}{a^2bc}\cdot\frac{(c+a)^2(s-c)(s-a)}{abc^2}\le 1.$

However, as we know, say,

$\displaystyle \frac{(b+c)^2(s-b)(s-c)}{a^2bc}=\cos^2\frac{\beta -\gamma}{2},$

which reduces the required inequality to the trivial

$\displaystyle \cos^2\frac{\alpha -\beta}{2}\cdot\cos^2\frac{\beta -\gamma}{2}\cdot\cos^2\frac{\gamma -\alpha}{2}\le 1,$

with equality only when all three angles are equal.

Solution 4

Lemma

Let $AX,\;$ $BY,\;$ $CZ\;$ be concurrent cevians such that $\displaystyle\frac{XB}{XC}=\frac{z}{y},\;$ $\displaystyle\frac{YC}{YA}=\frac{x}{z},\;$ and $\displaystyle\frac{ZA}{ZB}=\frac{y}{x},\;$ where $x,y,z\gt 0.\;$ Then

$\displaystyle [XYZ]=\frac{2xyz}{(x+y)(y+z)(z+x)}\cdot S,$

where $S=[\Delta ABC].$

Indeed, we find that $\displaystyle\frac{XB}{BC}=\frac{z}{y+z},\;$ implying $\displaystyle XB=\frac{az}{y+z}.\;$ Similarly, $\displaystyle ZB=\frac{cx}{x+y}.\;$ Hence,

$\displaystyle\begin{align} 2[\Delta BXZ]&=\frac{az}{y+z}\cdot\frac{cx}{x+y}\cdot\sin B\\ &=\frac{z}{y+z}\cdot\frac{x}{x+y}\cdot ac\sin B\\ &=\frac{2xz}{(x+y)(y+z)}\cdot S. \end{align}$

It follows that $\displaystyle [\Delta BXZ]=\frac{xz}{(x+y)(y+z)}\cdot S.\;$ Similarly, $\displaystyle [\Delta AZY]=\frac{yz}{(x+y)(z+x)}\cdot S\;$ and $\displaystyle [\Delta CYX]=\frac{xy}{(y+z)(z+x)}\cdot S.\;$ Now

$\displaystyle\begin{align}[\Delta XYZ]&=S-\left([\Delta BXZ]+[\Delta AZY] + [\Delta CYX]\right)\\ &=\frac{2xyz}{(x+y)(y+z)(z+x)}\cdot S. \end{align}$

Back to the main proof, we apply the lemma for $X=M,\;$ $Y=N,\;$ $Z=P\;$ and $\displaystyle x=\frac{1}{s-a},\;$ $\displaystyle y=\frac{1}{s-b},\;$ $\displaystyle z=\frac{1}{s-c}\;$ to obtain

$\displaystyle [\Delta MNP]=\frac{2(s-a)(s-b)(s-c)}{abc}\cdot S.$

Similarly, for $X=D,\;$ $Y=E,\;$ $Z=F\;$ and $x=a,\;$ $y=b,\;$ $z=c\;$ we obtain

$\displaystyle [\Delta DEF]=\frac{2abc}{(a+b)(b+c)(c+a)}\cdot S.$

We need to prove that

$\displaystyle \frac{(s-a)(s-b)(s-c)}{abc}\le \frac{abc}{(a+b)(b+c)(c+a)}.$

By Heron's formula $\displaystyle (s-a)(s-b)(s-c)=\frac{S^2}{s};\;$ also, $abc=4RS\;$ and $\displaystyle s=4R\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}.$

But

$\displaystyle a+b=2R(\sin A+\sin B)=4R\cos\frac{C}{2}\cos\frac{A-B}{2},\\ \displaystyle b+c=2R(\sin B+\sin C)=4R\cos\frac{A}{2}\cos\frac{B-C}{2},\\ \displaystyle c+a=2R(\sin C+\sin A)=4R\cos\frac{B}{2}\cos\frac{C-A}{2}.$

Hence our inequality is equivalent to

$\displaystyle\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}\cos\frac{A-B}{2}\cos\frac{B-C}{2}\cos\frac{C-A}{2}\le\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$

which is obviously true.

Acknowledgment

The problem has been proposed by Leonard Giugiuc and Daniel Sitaru. The solution was published in Crux Mathematicorum, VOLUME 42, NO. 2 February / Février 2016, with the following editorial remark:

We received eleven submissions, of which nine were correct and complete. We present the solution by Sefket Arslanagic, slightly modified by the editor.

Thus, Solution 1 is by Sefket Arslanagic. I am grateful to Marian Dinca who communicated to me the problem, Sefket Arslanagic's solution, and a solution (Solution 2) of his own. Solution 3 is by yours truly. Solution 4 is by Leonard Giugiuc and Daniel Sitaru.

 

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