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Let the feet of angle bisectors of triangle $ABC$ are $X,Y$ and $Z$ . The circumcircle of triangle $XYZ$ cuts off three segment from lines $AB,BC $and $CA$, let it be $XX’, YY’, ZZ’$. Prove that at least one combination exists such that sum of two segment's length is equal to third segment’s length.

WLOG, I assumed

$YY’\geq XX’\geq ZZ’$, So we have to prove $YY’=XX’+ZZ’$ enter image description here

I tried it using coordinate geometry assuming $A,B,C$ to be $(x_i,y_i)$ for $i=1,2,3$ . Then we can find coordinates of $X,Y,Z$ and hence the equation of circumcircle of $\Delta XYZ$. And then intercept with three sides. But it will consume a whole to day If a human follows this method without computer help.

Can someone please help me finding a good solution for this problem?

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  • $\begingroup$ Maybe I did something wrong, by the intersecting secant theorem we have $$CY\cdot YY'= CX \cdot XX'$$ and similar relation for the other two pairs of secant. Assume the triangle is equilater, then perpendicular bisector are heights and so the are the height of the triangle $\triangle CXY$, and so such triangle is isosceles, thus $CX=CY$ and so $$XX'=YY'$$. By repeating the process for $YY'$ and $ZZ'$ you obtain $$XX'=YY'=ZZ'$$ and the theorem is true only they are all zero. I can't spot my mistake. Is there some condition on the triangle? $\endgroup$
    – Marco
    Commented Mar 3 at 12:02
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    $\begingroup$ @Marco by intersecting secant theorem, $$CY\cdot CY’= CX\cdot CX’$$ $\endgroup$
    – Leibniz-Z
    Commented Mar 3 at 12:46
  • $\begingroup$ Right, but if $CX=CY$ (that happen when the triangle is equilateral) then you have $$CX(CX+XX')=CY(CY+YY') \rightarrow XX'=YY'$$ $\endgroup$
    – Marco
    Commented Mar 3 at 13:08
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    $\begingroup$ @Marco yes all are zero in equilateral triangle. Since the given circle will be incircle in that case. $\endgroup$
    – Leibniz-Z
    Commented Mar 3 at 13:12
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    $\begingroup$ What is the source? Most euclidean geometry questions like this are from a contest. $\endgroup$
    – D S
    Commented Mar 3 at 15:12

2 Answers 2

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For the sake of simplicity, let's assume $AB=c, AC=b, AB=c, XX'=x, YY'=y,$ and $ZZ'=z$. Just by using the properties of the power of a point and angle bisectors, we will have:

$$\frac{ac}{b+c}(\frac{ac}{b+c}-x)=\frac{ca}{a+b}(\frac{ca}{a+b}+z), \\ \frac{bc}{a+b}(\frac{bc}{a+b}-z)=\frac{bc}{a+c}(\frac{bc}{a+c}-y), \\ \frac{ab}{a+c}(\frac{ab}{a+c}+y)=\frac{ab}{b+c}(\frac{ab}{b+c}+x),$$

or equivalently,

$$\frac{1}{b+c}(\frac{ac}{b+c}-x)=\frac{1}{a+b}(\frac{ca}{a+b}+z), \\ \frac{1}{a+b}(\frac{bc}{a+b}-z)=\frac{1}{a+c}(\frac{bc}{a+c}-y), \\ \frac{1}{a+c}(\frac{ab}{a+c}+y)=\frac{1}{b+c}(\frac{ab}{b+c}+x).$$

By solving this $3-$variable system of equations, we will get:

$$2x=\frac{a(c-b)}{b+c}+\frac{b(b+c)}{a+c}-\frac{c(b+c)}{a+b}, \\2z=\frac{c(b-a)}{a+b}+\frac{a(a+b)}{b+c}-\frac{b(a+b)}{a+c}, \\ 2y=\frac{b(c-a)}{a+c}+\frac{a(a+c)}{b+c}-\frac{c(a+c)}{a+b}.$$

Now, it's almost obvious that we have $2x+2z=2y.$

We are done.


Note $1$: To clarify the initial relations, we have used the fact that, for example, $BX=\frac{ac}{b+c}$ and $BX(BX-XX')=BZ(BZ+ZZ').$

Note $2$: The $3-$variable system of linear equations can be easily solved by hand.

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  • $\begingroup$ It seems this computational approach is unavoidable, as the best I can show is that AX',BY',CZ' are concurrent. $\endgroup$
    – D S
    Commented Mar 3 at 15:24
  • $\begingroup$ @DS Usually, these contest problems and their claims don't come out of nowhere. So, it is likely for the official solution to be non-computational. However, if this is a contest and the OP has to get the scores, the best way is not always the shortest or the most elegant one. If this is just a challenge or practice, the OP can (or should !) spend a lot of time not only to find the nicer solution but also to gain the some geometric intuition through the thinking process. I hope you (or someone else) will come up with an elegant approach! $\endgroup$ Commented Mar 3 at 15:41
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I will outline an another approach that is useful to remember for Euclidean/Olympiad geometry problems. A simple application of Theorem of Sines gives us: $$XX' = 2r\sin\left(\frac\alpha 2+\gamma\right)=2r\sin\left(\frac\alpha 2+\beta\right)=r\left(\sin\left(\frac\alpha 2+\gamma\right)+\sin\left(\frac\alpha 2+\beta\right)\right) = $$ $$=2r\sin\frac{\pi}{2}\cos\frac{\beta-\gamma}{2} = 2r\cos\frac{\beta-\gamma}{2}.$$

From here, it suffices to prove these $\cos\frac{\beta-\gamma}{2}, \cos\frac{\gamma-\alpha}{2}, \cos\frac{\alpha-\beta}{2}$ satisfy your condition. Then you can brute force this however way you want - I personally would start out by finding: $$\sin\frac{\alpha}{2} = \sqrt{\frac{1-\cos\alpha}{2}} = \sqrt{\dfrac{(a-b+c)(a+b-c)}{4bc}} = \sqrt{\frac{(p-b)(p-c)}{bc}}$$ and then you get: $$\cos\frac{\alpha}{2} = \sqrt{\dfrac{p(p-a)}{bc}}$$ and so on.

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