All-Russian Olympiad 2010 11 grade P-6

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Ovchinnikov Denis

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Ovchinnikov Denis

#1 h Sep 10, 2010, 4:52 PM • 2 Y

Y by Adventure10, Mango247

Could the four centers of the circles inscribed into the faces of a tetrahedron be coplanar?

(vertexes of tetrahedron not coplanar)

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#2 h Sep 11, 2010, 8:32 AM • 2 Y

Y by Adventure10, Mango247

$ \mbox{In tetrahedron }A_1A_2A_3A_4\mbox{ denote by: }a_{ij}=|A_iA_j| \mbox{ and if }\\ I_k\mbox{-is the incenter of }\triangle{A_iA_jA_h} \mbox{ for: }\{i;j;h;k\}=\{1;2;3;4\},\\ \mbox{then in barycentric coordinate we have: }\\ I_1(0;a_{34};a_{24};a_{23}); I_2(a_{34};0;a_{14};a_{13}); I_3(a_{24};a_{14};0;a_{12}); I_4(a_{23};a_{13};a_{12};0)\\ \mbox{and}\\ \mbox{The points }I_1;I_2;I_3;I_4\mbox{-is coplanar}\Leftrightarrow \left|\begin{array}{cccc} 0&a_{34}&a_{24}&a_{23}\\ a_{34}&0&a_{14}&a_{13}\\ a_{24}&a_{14}&0&a_{12}\\ a_{23}&a_{13}&a_{12}&0 \end{array}\right|=0\Leftrightarrow\\ \Leftrightarrow a_{12}^2a_{34}^2+a_{13}^2a_{24}^2+a_{14}^2a_{23}^2= 2.\left(a_{12}a_{13}a_{24}a_{34}+a_{12}a_{14}a_{23}a_{34}+a_{13}a_{14}a_{23}a_{24}\right).\ \ (1)\\ \mbox{But:}\\ \left|a_{13}a_{24}-a_{14}a_{23}\right|

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1276 posts

va2010

#3 h Aug 18, 2016, 2:53 AM • 1 Y

Y by Adventure10

Use $4$ -dimensional barycentric coordinates on $ABCD$ . Where $I_{ABC}$ denotes the incenter of $ABC$ , and so on, we obtain the coordinates

$\begin{align*} I_{ABC} &= (BC : AC : AB : 0) \\ I_{ABD} &= (BD : AD : 0 : AB) \\ I_{ACD} &= (CD : 0 : AD : AC) \\ I_{BCD} &= (0 : BC : BD : CD) \end{align*}$
Coplanarity is the condition that the above determinant is zero. This rearranges to $\operatorname{det} X = 0$ , where
$X = \left( {\begin{array}{*{20}c} 0 & AB & AC & BC \\ AB & 0 & AD & BD \\ AC & AD & 0 & CD \\ BC & BD & CD & 0 \\ \end{array} } \right)$

The proof that this determinant is not zero is rather challenging. First of all, from here we see that the determinant being equal to zero is equivalent to the statement that $\sqrt{(AB)(CD)} + \sqrt{BC(AD)} = \sqrt{(AC)(BD)},$ more or less (symmetry is clear). The proof that this does not occur is quite pleasant: observe that
$\sqrt{(AB)(CD)} + \sqrt{(BC)(AD)} \ge \sqrt{(AB + BC)(AD + CD)} > AC$ and
$\sqrt{(AB)(CD)} + \sqrt{(BC)(AD)} \ge \sqrt{(AB + AD)(CD + BC)} > BD$ so the result is obvious.

This post has been edited 1 time. Last edited by va2010, Aug 18, 2016, 2:54 AM

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yayups

#4 h Nov 8, 2018, 1:25 AM • 2 Y

Y by Adventure10, Mango247

Hmm it appears I have found a slightly simpler way to finish. The answer is $\boxed{\mathrm{no}}$ . Again, use the 4D bary, and if they were coplanar, we have
$\begin{vmatrix}0 & AB & AC & BC \\ AB & 0 & AD & BD \\ AC & AD & 0 & CD \\ BC & BD & CD & 0 \\ \end{vmatrix}=0.$ This can be expanded to
$(AB\cdot CD-AC\cdot AB)^2 + (AB\cdot CD-AD\cdot BC^2)^2 + (AC\cdot BD-AD\cdot BC)^2 - (AB^2\cdot CD^2+AC^2\cdot BD^2 + AD^2\cdot BC^2)=0.$ Define $a=AB\cdot CD$ , $b=AC\cdot BD$ , and $c=AD\cdot BC$ . Ptolemy's inequality tells us that $a,b,c$ satisfy pairwise triangle inequalities. Therefore, we can use the age old Ravi substitution to write $a=y+z$ , $b=x+z$ , $c=x+y$ for $x,y,z>0$ . Thus, we have
$(x-y)^2+(y-z)^2+(z-x)^2 = (x+y)^2 + (y+z)^2 + (z+x)^2.$ But if $p,q\ge 0$ , then $|p-q|<|p+q|$ , so $(p-q)^2<(p+q)^2$ . Thus, the LHS of the above equation is strictly smaller than that of the RHS, so we have the desired contradiction.

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#5 h Jan 15, 2020, 7:43 AM • 1 Y

Y by Adventure10

Here's yet another finish (by compressing the tetrahedron). What a nice problem!

Ovchinnikov Denis wrote:

Could the four centers of the circles inscribed into the faces of a tetrahedron be coplanar?

(vertexes of tetrahedron not coplanar)

We claim that the answer is no. To prove this, we extend barycentric coordinates to $3$ Dimension as follows:

Fix a reference tetrahedron $ABCD$ with sides $BC=a, CA=b, AC=b, DA=d, DB=e, DC=f.$ Then the coordinates of any point $P$ are defined by
$P:=\frac{1}{[ABCD]}\Big( [PBCD],[PCDA],[PDAB],[PABC] \Big)$ where the areas are signed.

Hence, if $\mathcal{I} (\triangle)$ denotes the incenter of triangle $\triangle,$ then it is easy to get (using the areal definition):
$\begin{align*} \mathcal{I} (ABC)=(a:b:c:0) \end{align*}$ where the coordinates are un-homogenized. We get similar coordinates for the incenters of the other faces. Thus, these being coplanar is equivalent to
$\begin{align*} \det \begin{bmatrix} 0 & f & e & a \\ f & 0 & d & b \\ e & d & 0 & c \\ a & b & c & 0 \end{bmatrix}=0 \end{align*}$ Let $u=\sqrt{ad}, v=\sqrt{be}, w=\sqrt{cf}.$
The determinant evaluates to
$-u^4-v^4-w^4+2u^2v^2+2v^2w^2+2w^2u^2=0$ This implies
$(u+v+w)(u+v-w)(u-v+w)(-u+v+w)=0$ Assume without loss of generality that $u=v+w.$ Squaring implies $ad=be+cf+2\sqrt{becf}>be+cf.$

Now comes the trick. Compress the tetrahedron to a quadrilateral $ACDB$ (notice the order). This order of vertices ensures that $b,c,e,f$ are the sides and $a,d$ are the diagonals. So by Ptolemy's inequality, $be+cf>ad,$ and so we have a contradiction. $\blacksquare$

Note: The compression here is not exactly a compression. What we do is create a quadrilateral with corresponsing sides and diagonals equal to the corresponding sides of the tetrahedron. Such a quadrilateral exists by the triangle inequality on all the faces.

This post has been edited 1 time. Last edited by Wizard_32, Jan 31, 2020, 10:16 AM

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#6 h Jan 15, 2020, 9:53 AM • 1 Y

Y by Adventure10

$\mbox{In tetrahedron }A_1A_2A_3A_4\mbox{ denote by: }a_{ij}=|A_iA_j| \mbox{ and if }\\ I_k\mbox{-is the incenter of }\triangle{A_iA_jA_h} \mbox{ for: }\{i;j;h;k\}=\{1;2;3;4\},\\ \mbox{then in barycentric coordinate we have: }\\ I_1(0;a_{34};a_{24};a_{23}); I_2(a_{34};0;a_{14};a_{13}); I_3(a_{24};a_{14};0;a_{12}); I_4(a_{23};a_{13};a_{12};0)\\ \mbox{and}\\ \mbox{The points }I_1;I_2;I_3;I_4\mbox{-is coplanar}\Leftrightarrow \left|\begin{array}{cccc} 0&a_{34}&a_{24}&a_{23}\\ a_{34}&0&a_{14}&a_{13}\\ a_{24}&a_{14}&0&a_{12}\\ a_{23}&a_{13}&a_{12}&0 \end{array}\right|=0\Leftrightarrow\\ \Leftrightarrow a_{12}^2a_{34}^2+a_{13}^2a_{24}^2+a_{14}^2a_{23}^2= 2.\left(a_{12}a_{13}a_{24}a_{34}+a_{12}a_{14}a_{23}a_{34}+a_{13}a_{14}a_{23}a_{24}\right).\ \ (1)\\ \mbox{But:}\\ \left|a_{13}a_{24}-a_{14}a_{23}\right|$

I guess what he meant :-D

:-D

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algebra_star1234

2467 posts

algebra_star1234

#7 h Jun 13, 2020, 11:23 PM

Y by

We will use 3D barycentric coordinates. The incenter of $BCD$ is $I_A = (0:CD:BD:BC)$ . The incenter of $ACD$ is $I_B = (CD:0:AD:AC)$ . The incenter of $ABD$ is $I_C = (BD:AD:0:AB)$ . The incenter of $ABC$ is $I_D = (BC:AC:AB:0)$ . The points are coplanar if
$\begin{vmatrix} 0 & CD & BD & BC \\ CD & 0 & AD & AC \\ BD & AD & 0 & AB \\ BC & AC & AB & 0 \end{vmatrix} = 0$ Therefore,
$\begin{align*} (CD \cdot AB)^2 + (BD \cdot AC)^2 + (BC \cdot AD)^2 - 2(CD \cdot AB)(BD\cdot AC) - 2(CD \cdot AB)(BC \cdot AD)&\\ - 2(BD \cdot AC)(BC \cdot AD)=0 \end{align*}$ Let $CD \cdot AB = x$ , $BD \cdot AC = y$ , and $BC \cdot AD = z$ . Then, we have
$x^2+y^2+z^2 -2xy-2yz-2zx = 0$ Writing this as a quadratic in $x$ , we have either $x = y+z + 2\sqrt{yz}$ or $x = y+z - 2\sqrt{yz}$ . In the second case, we either have $y = x+z + 2\sqrt{xz}$ or $z = x+y + 2\sqrt{xy}$ . However, none of these can happen because we can show that $x+y \ge z$ and similarly for the other permutations. In fact, consider the projection of the tetrahedron onto the plane parallel to $BC$ and $AD$ . Let $x'$ and $y'$ be the values of $x$ and $y$ in the projection. Note that $x \ge x'$ and $y \ge y'$ . Then,
$x+y \ge x' + y' \ge z$ by Ptolemy's inequality. Thus, we are done.

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#8 h Feb 13, 2024, 4:41 PM

Y by

Wow, no synthetic solutions posted. Here is a one (diagrams from official):
We take 2 cases :
CASE 1: $I_AI_BI_CI_D$ is convex quadrilateral.
The idea is to introduce the midpoints of $AD,AB,CD,CB$ , call them $N,M,K,L$ .

https://cdn.artofproblemsolving.com/attachments/5/7/2e7810d53c9c21cf2d9a065ff5366aec203d40.png

$\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad \quad \quad$ Draw the line $\mathbf{L}$ parallel through $B$ to $AC$ , Comparing the distance of $I_D$ from $\mathbf{L}$ and $AC$ we find that $I_D$ is below the midline $LM$ . So we have that $I_B,I_D$ and $I_A,I_C$ lie on opposite sides of plane and hence can't intersect contradiction.

https://cdn.artofproblemsolving.com/attachments/8/6/8d2bc2aa6ee06a220672408cbe1f7139eefbeb.png

CASE 2: It is concave
This happens when $I_A$ lies inside $I_BI_CI_D$ but however, $I_A \in BCD$ while the other lies above contradiction!.
Done!

This post has been edited 10 times. Last edited by math_comb01, Nov 24, 2024, 4:44 PM

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