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enter image description hereSuppose you are given an acute triangle $XYZ$ with the following properties:

At $\angle XZY$, the $\angle$ bisector is drawn and extended all the way to $XY$. Lets call the point where it intersects $A$. From $\angle ZXY$ we draw a line to $ZY$ s.t. it cuts $ZY$ into two equal pieces. i.e. the point that intersects $ZY$ is the midpoint call it $B$. And lastly, from $\angle ZYX$ we drop the altitude onto $XZ$. Lets call the point it intersects $C$. Now this triangle was constructed s.t. $ZA \cap XB \cap YC = P$, a point. Suppose we let $\angle XZY = \phi$. How would we prove that $\angle XZY > 45$ degrees for this specific kind of triangle? Obviously my first thought was to suppose that it was $\leq 45$ degrees, and then break it up into two cases. Even then I am unsure. The help would be appreciated!

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marked as duplicate by Blue, Hakim, Mark Bennet, John, Ali Caglayan Nov 12 '14 at 0:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ A picture is worth a thousand words. $\endgroup$ – Edward Jiang Nov 11 '14 at 18:05
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    $\begingroup$ There you go :) $\endgroup$ – Nick Freeman Nov 11 '14 at 18:45
  • $\begingroup$ Hmmm...I give that picture a score of 500 words, because (i) it looks like triangle $XZY$ is isosceles, (ii) that's the biggest right angle I've seen in a long while! $\endgroup$ – TonyK Nov 11 '14 at 19:02
  • $\begingroup$ Its just a rough sketch. I hope it is useful still! $\endgroup$ – Nick Freeman Nov 11 '14 at 19:08
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Let's choose some coordinates, w.l.o.g.:

$$X=(0,0)\qquad Y=(1,0)\qquad Z=(t,u)$$

Then you have

$$XB:ux=(t+1)y\qquad YC:tx+uy=t$$

which intersect in

$$P=(t^2+t,tu)/(u^2+t^2+t)$$

Now the (quadratic) condition $\angle XZP=\angle PZY$ can be rephrased as

\begin{align*} \frac{\langle X-Z,P-Z\rangle}{\lVert X-Z\rVert\cdot\lVert P-Z\rVert} &= \frac{\langle P-Z,Y-Z\rangle}{\lVert P-Z\rVert\cdot\lVert Y-Z\rVert} \\ \langle X-Z,P-Z\rangle^2\cdot\lVert Y-Z\rVert^2 &= \langle P-Z,Y-Z\rangle^2\cdot\lVert X-Z\rVert^2 \\ (u^2+t^2-t)^2(u^2+t^2-2t+1) &= \left(\frac{u^4+2u^2t^2+t^4-t^3-tu^2-t^2+t}{u^2+t^2+t}\right)^2(t^2+u^2) \\ \end{align*}

$$u^2 (t^6 + 3t^4u^2 + 3t^2u^4 + u^6 - 2t^5 - 4t^3u^2 - 2tu^4 + 2t^3 - t^2)=0$$

So if $u\neq 0$, then the second term characterizes all possible locations of the third corner. That's an algebraic curve of degree $6$.

You have to show that for $0<t<1$ this only results in real solutions which lie within the circle whose inscribed angle would be $45°$. Next I'd plot that algebraic curve:

Curve

As you can see, the blue curve stays within the red circles, so the angle is larger than $45°$. And the curve stays outside the orange circle, so the triangle is acute in all these cases.

One possible triangle satisfying your requirements would be the following example:

Triangle

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  • $\begingroup$ i understood the answer $\endgroup$ – lokesh sangabattula Nov 18 '14 at 16:16

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