In a triangle $ABC$, let $A_{1}$, $B_{1}$, $C_{1}$ be the points where the excircles touch the sides $BC$, $CA$ and $AB$ respectively. Prove that $A A_{1}$, $B B_{1}$ and $C C_{1}$ are the sidelenghts of a triangle. Proposed by L. Emelyanov
Problem
Source: Tuymaada Mathematical Olympiad 2005, Day 1, Problem 4
Tags: inequalities, geometry, trigonometry, 3D geometry, tetrahedron, triangle inequality, geometry unsolved
26.11.2005 16:30
Namdung mentioned a "very simple geometric proof" in http://www.mathlinks.ro/Forum/viewtopic.php?t=46088 . However, apparently nobody else has seen this proof until now, so let me post my proof, neither simple nor geometric, but actually showing some stronger inequalities. Note that it is rather long since I tried to avoid massive computations, but it is rather a typical inequality proof than a geometry proof, as you will surely see. First I rename some points: Problem. Let ABC be a triangle, and let its A-excircle, its B-excircle and its C-excircle touch its sides BC, CA and AB at the points X, Y and Z, respectively. Prove that the segments AX, BY and CZ are the sidelengths of a triangle. Solution. We will show that $BY^2+CZ^2>AX^2$, $CZ^2+AX^2>BY^2$ and $AX^2+BY^2>CZ^2$. This will yield $\left(BY+CZ\right)^2=BY^2+CZ^2+2\cdot BY\cdot CZ>BY^2+CZ^2>AX^2$, so that BY + CZ > AX, and similarly CZ + AX > BY and AX + BY > CZ. Thus, the segments AX, BY and CZ are the sidelengths of a triangle, and moreover, since $BY^2+CZ^2>AX^2$, $CZ^2+AX^2>BY^2$ and $AX^2+BY^2>CZ^2$, this triangle must be acute-angled, so we thus show even more than the problem asks us for! So, it remains to prove the inequalities $BY^2+CZ^2>AX^2$, $CZ^2+AX^2>BY^2$ and $AX^2+BY^2>CZ^2$. We will only prove $BY^2+CZ^2>AX^2$, since the other two inequalities will then follow by cyclic translation. Let $s =\frac{a+b+c}{2}$ be the semiperimeter of triangle ABC. Remember the half-angle formula $\cos^2\frac{B}{2}=\frac{s\left(s-b\right)}{ca}$. Since $s =\frac{a+b+c}{2}$, we have $s-b=\frac{a+b+c}{2}-b=\frac{c+a-b}{2}$, and thus, this formula becomes $\cos^2\frac{B}{2}=\frac{\frac{a+b+c}{2}\cdot\frac{c+a-b}{2}}{ca}=\frac{\left(c+a+b\right)\left(c+a-b\right)}{4ca}=\frac{\left(c+a\right)^2-b^2}{4ca}$, so that $\left(c+a\right)^2-b^2=4ca\cos^2\frac{B}{2}$. Now, applying this formula to the triangle ABX, we get $\left(AB+BX\right)^2-AX^2=4\cdot AB\cdot BX\cdot\cos^2\frac{\measuredangle ABX}{2}$. Now, we have AB = c, and since X is the point where the A-excircle of triangle ABC touches its side BC, we have BX = s - c. Thus, AB + BX = c + (s - c) = s. Finally, < ABX = B. Hence, this equation becomes $s^2-AX^2=4\cdot c\cdot\left(s-c\right)\cdot\cos^2\frac{B}{2}$. But since $\cos^2\frac{B}{2}=\frac{s\left(s-b\right)}{ca}$, this transforms into $s^2-AX^2=4\cdot c\cdot\left(s-c\right)\cdot\frac{s\left(s-b\right)}{ca}=\frac{4s\left(s-b\right)\left(s-c\right)}{a}$, and thus $AX^2=s^2-\frac{4s\left(s-b\right)\left(s-c\right)}{a}=s\left(s-\frac{4\left(s-b\right)\left(s-c\right)}{a}\right)$. Now, denote s - a = x, s - b = y and s - c = z (note that these numbers x, y, z are then positive). Then, y + z = (s - b) + (s - c) = 2s - (b + c) = (a + b + c) - (b + c) = a, and x + y + z = x + (y + z) = (s - a) + a = s. Thus, this becomes $AX^2=\left(x+y+z\right)\left(x+y+z-\frac{4yz}{y+z}\right)$. Similarly, $BY^2=\left(x+y+z\right)\left(x+y+z-\frac{4zx}{z+x}\right)$; $CZ^2=\left(x+y+z\right)\left(x+y+z-\frac{4xy}{x+y}\right)$. Hence, the inequality we want to prove, $BY^2+CZ^2>AX^2$, is equivalent to $\left(x+y+z\right)\left(x+y+z-\frac{4zx}{z+x}\right)+\left(x+y+z\right)\left(x+y+z-\frac{4xy}{x+y}\right)>\left(x+y+z\right)\left(x+y+z-\frac{4yz}{y+z}\right)$. Upon division by x + y + z, this becomes $\left(x+y+z-\frac{4zx}{z+x}\right)+\left(x+y+z-\frac{4xy}{x+y}\right)>x+y+z-\frac{4yz}{y+z}$. Equivalently, $x+y+z+\frac{4yz}{y+z}>\frac{4zx}{z+x}+\frac{4xy}{x+y}$. Now, we can directly prove this inequality by expanding, but this is a bit uncreative and requires computations. Let's prove this inequality in a slightly different way, by showing a stronger symmetric inequality. In fact, first we rewrite this inequality by adding $\frac{4yz}{y+z}$ to both sides: (1) $x+y+z+\frac{8yz}{y+z}>4\left(\frac{yz}{y+z}+\frac{zx}{z+x}+\frac{xy}{x+y}\right)$. Now, since (z + x) (x + y) = x (x + y + z) + yz > x (x + y + z), we have $\frac{8yz\left(z+x\right)\left(x+y\right)}{y+z}>\frac{8yzx\left(x+y+z\right)}{y+z}=8\frac{xyz\left(x+y+z\right)}{y+z}$, so that $\frac{8yz}{y+z}>8\frac{xyz\left(x+y+z\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}$. Thus, the inequality (1) would immediately follow from the following symmetric inequality: (2) $x+y+z+8\frac{xyz\left(x+y+z\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\geq 4\left(\frac{yz}{y+z}+\frac{zx}{z+x}+\frac{xy}{x+y}\right)$. Now, though this inequality be destroyed by expanding, it's quite sharp. Here is a halfways "synthetic" proof of this inequality: By subtracting $4\frac{xyz\left(x+y+z\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}$ from both sides of the inequality (2), we bring it into the form $x+y+z+4\frac{xyz\left(x+y+z\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\geq 4\left(\frac{yz}{y+z}+\frac{zx}{z+x}+\frac{xy}{x+y}-\frac{xyz\left(x+y+z\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\right)$. But this can be proven as follows: $x+y+z+4\frac{xyz\left(x+y+z\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}=\left(x+y+z\right)\cdot\left(1+\frac{4xyz}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\right)$ $=\left(x+y+z\right)\cdot\frac{4xyz+\left(y+z\right)\left(z+x\right)\left(x+y\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}$ $=\left(x+y+z\right)\cdot\frac{x\left(y+z\right)^2+y\left(z+x\right)^2+z\left(x+y\right)^2}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}$ $=\frac{\left(x+y+z\right)\cdot\left(x\left(y+z\right)^2+y\left(z+x\right)^2+z\left(x+y\right)^2\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}$ $\geq\frac{\left(\sqrt{x\cdot x\left(y+z\right)^2}+\sqrt{y\cdot y\left(z+x\right)^2}+\sqrt{z\cdot z\left(x+y\right)^2}\right)^2}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}$ (by Cauchy-Schwarz) $=\frac{\left(x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)\right)^2}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}=\frac{4\left(yz+zx+xy\right)^2}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}$ $=4\left(yz\frac{yz+zx+xy}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}+zx\frac{yz+zx+xy}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}+xy\frac{yz+zx+xy}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\right)$ $=4\left(\left(\frac{yz}{y+z}-\frac{x^2yz}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\right)+\left(\frac{zx}{z+x}-\frac{y^2zx}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\right)+\left(\frac{xy}{x+y}-\frac{z^2xy}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\right)\right)$ $=4\left(\frac{yz}{y+z}+\frac{zx}{z+x}+\frac{xy}{x+y}-\frac{x^2yz+y^2zx+z^2xy}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\right)$ $=4\left(\frac{yz}{y+z}+\frac{zx}{z+x}+\frac{xy}{x+y}-\frac{xyz\left(x+y+z\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\right)$. And (2) is proven, so the problem is solved. Darij
26.11.2005 20:37
In http://www.mathlinks.ro/Forum/topic-59415.html Darij elegantly convinced us that Medians triangle is always possible. Nagelians are isotomics of Gergonnenians, so they should always be on the 'right' side of Medians and thus they should not struggle for triangle existence. Thank you. M.T.
25.12.2005 18:11
I don't understand what you are saying
25.12.2005 21:27
Medians of a triangle can form a triangle of their own, they have enough length in them. If this is true (Darij gave a convincing argument, so it is not iffy as it was before he delivered the proof), then cevians of Nagel point are even 'more triangulable' then medians, if you consider position of Nagel point w.r.t. Centroid and Gergonne point. Salute and thank you. M.T.
25.12.2005 23:00
I don't call that a proof, armpist...
26.12.2005 02:10
Arne wrote: I don't call that a proof, armpist... It is perfectly all right, it is not your fault, it happens with my proofs from time to time, they have that deceiving first impression. Yet, I suspect that you did not spend enough time trying to fathom the underlying original and fruitful idea. You can and probably should for your own good get back to it after the Holidays, when you will be able to concentrate exclusively on it. Happy New Year to All ! Thank you. M.T.
26.12.2005 03:10
Oh right, sure
17.05.2011 18:42
Dear Perfect_radio where is "Tuymaada"?
17.05.2011 18:49
take a look at this.
23.06.2011 03:38
Consider the triangle $A'B'C'$ whose medial triangle is $ABC$ and let $D$, $E$, $F$ be the tangency points of the incircle of $ABC$ with $BC$, $CA$, $AB$ respectively. Then $A'D=AA_1$, $B'E=BB_1$ and $C'F=CC_1$. Assuming WLOG that $\angle A\le90^\circ$ fold the triangles $ABC'$ and $ACB'$ along the sides $AB$ and $AC$ until $B'$ and $C'$ coincide in a point $P$. If $\angle A<90^\circ$ there are two possibilities for $P$: one above the plane of $ABC$ and one below, mirror images, so it does not matter which one we choose. In any case the triangle $PEF$ is formed and $PE=BB_1$, $PF=CC_1$. If $\angle A=90^\circ$ then $P$ is unique and it is in the plane of $ABC$. By triangle inequality $PE+EF>PF$ and $PF+EF>PE$, but $EF$ is a chord of the incircle so $EF$ is smaller than an altitude of $ABC$, and so smaller than $AA_1$, hence $PE+AA_1>PF$ and $PF+AA_1>PE$, i.e., $BB_1+AA_1>CC_1$ and $CC_1+AA_1>BB_1$. Folding along the other acute angle we obtain the remaining inequality. Observe that if the triangle is acute then we can fold the three triangles and obtain a nice tetrahedron. If the triangle is rectangle we obtain a degenerate tetrahedron when folding along the right angle. When the triangle has an obtuse angle then it is impossible to form a tetrahedron by folding the three triangles around $ABC$.