6
$\begingroup$

Prove that the lines through $A$ and the incenter of $\Delta ABC $, through $B$ and the circumcenter of $\Delta ABC$, and through $C$ and the orthocenter of $\Delta ABC $ are concurrent if and only if $\cos^{2} A =\cos B \cdot \cos C $.

My attempt:enter image description here

By Ceva's Theorem we have that $\cfrac {AP \cdot CQ \cdot BR}{PC \cdot BQ \cdot AR}=1$,which can be rearranged in the form $\cos^2 A=\left(\cfrac {AP \cdot CQ \cdot BR}{PC \cdot BQ \cdot AC}\right)^2$, from that I've tried to solve the problem working on these two directions:

1)Show that $\left(\cfrac {AP \cdot CQ \cdot BR}{PC \cdot BQ \cdot AC}\right)^2=\cos B \cdot \cos C$

2) Since we have also that $\cos^2 A = \left( \cfrac {AR}{AC} \right)^2 $, we have to show that $\cos B \cdot \cos C = \left ( \cfrac {AR}{AC} \right)^2 $

I've worked most on the second line since it seems simpler and that's what I was able to do so far:

Given that $ \cos B = \cfrac {BR}{RC} $ and $\cos C =\cfrac {BC^2 +AC^2-AB^2}{2AC \cdot BC} $, I have in the end (after some algebraic manipulations):

$AC^2 +CB^2-AB^2 =\cfrac {2AR ^2 \cdot BC^2}{AC \cdot BR}$ and that's where my tombstone is. I don't know how to simplify this any further, I don't know if it is even worth to simplify it given that this might be the wrong path to take...

I know that I am not coming to a solution since I am not using the fact that $BP$ passes through the circumcenter $K$ of $\Delta ABC$ and that obviously is a key point in solving the problem but I don't know how to use this information.

Edit: I've tried the following: Let $J$ be the point of intersection of lines $BP$ (this line passes through the circumenter $K$) and line $CO$ (where $O$ is the orthocentre ) ,so now what i've to do is to prove that $A,J,Q$ are collinear,i.e. I have to prove that $$\cfrac{AP \cdot CQ \cdot BJ}{BQ \cdot AC \cdot PJ}=1 \tag 1$$ . Applying the Angle Bisector Theorem to $A$ I find $\cfrac{CQ}{BQ}=\cfrac { AC}{AB}$ from which i have got the following : $$ \cos^2 A = \left ( \cfrac {PJ \cdot BQ \cdot CQ}{BJ \cdot AP} \right)^2 $$ Since we know that $\cos B = \cfrac{BR}{BC}$ so i have to prove $$ \cos C = \left (\cfrac { PJ \cdot BQ \cdot CQ}{BJ \cdot AP} \right)^2 \cdot \cfrac {BC}{BR}$$ but so far i was unable to do that.

Any hint ,solution is appreciated.

*Only geometrical methods,please.

$\endgroup$
  • $\begingroup$ I think if you can prove triangle is equilateral then this is simply proved. $\endgroup$ – Archis Welankar Nov 6 '15 at 13:22
  • $\begingroup$ @ArchisWelankar If the drawing is accurate, then it need not be equilateral... $\endgroup$ – rogerl Nov 6 '15 at 13:30
  • $\begingroup$ yes ,it doesn't have to be equilateral. $\endgroup$ – Nameless Nov 6 '15 at 13:39
  • $\begingroup$ I am just giving an option to prove they are. there are many ways like determinant form if we know three points. $\endgroup$ – Archis Welankar Nov 6 '15 at 13:57
  • $\begingroup$ @ Archis Welankar I should add that the only methods i am allowed to use are purely geometrical ones. $\endgroup$ – Nameless Nov 6 '15 at 14:26
1
$\begingroup$

I'll be using the usual standard notation for the elements of $\triangle ABC$.

To solve this problem we need only to find the ratios ${QB\over QC},{PA\over PC} $ and ${RA\over RB} $ in terms of the elements.

We simply apply the Angle Bisector Theorem in $\triangle ABC$ to get our first ratio: $${QB\over QC}={c\over b}\tag{i}$$

Now just apply simple trigonometry in the triangles $\triangle RBC$ and $\triangle RAC$ to get the second ratio: $${RA\over RB}={b\cos A\over a\cos B}\tag{ii}$$

For the third ratio you have to do some work. Apply the Law of sines in the triangles $\triangle BPA$ and $\triangle BPC$ and obtain the relation: $\dfrac{PA}{PC}=\dfrac{\sin C}{\sin A}\cdot\dfrac{\sin \angle ABP}{\sin \angle CBP}\tag{iii}$

Then observe that $2R\cos\angle ABP=c$ and $2R\cos\angle CBP=a$. This will finally provide us the third ratio : $${PA\over PC}={\sin 2C\over \sin 2A} \tag{iv} $$

Now just apply Ceva's thoerem and you will directly obtain the result.

NOTE: You may need to use the formula $2R\sin X=x$ where $X\in\{A,B,C\}$ and $x\in\{a,b,c\}$

EDIT: Since $2R\cos\angle ABP=c$ and $2R\cos\angle CBP=a$ we have $${\sin\angle ABP\over \sin\angle CBP}={\sqrt{4R^2-c^2}\over \sqrt{4R^2-a^2}}={\cos C\over\cos A }\tag{v}$$

Apply (v) in (iii) to get (iv)

$\endgroup$
  • 1
    $\begingroup$ I am not quite sure of how we got that $\cfrac{PA}{PC}=\cfrac{ \sin 2C }{ \sin 2A}$,could you elaborate a bit ? $\endgroup$ – Nameless Nov 7 '15 at 8:18
  • $\begingroup$ @Nameless check out the edit. $\endgroup$ – G-man Nov 7 '15 at 14:38
  • $\begingroup$ One last question:I see that $\cfrac {\sin \angle ABP }{ \sin \angle CBP}= \cfrac { \cos C }{\cos A}$ ,by noticing $ \sin \angle ABP = \cos (90-ABP)= \cos C $ and same goes for $ \sin \angle CBP$, but when i look at your (v) i have the impression that you derived the above fact from a different point of view,is that the case ?(That's indeed my last question,thanks again G-man for your time) $\endgroup$ – Nameless Nov 7 '15 at 17:31
  • $\begingroup$ @Nameless yeah I somehow totally missed such an obvious point, the angles were complementary all the time! I applied the pythagorean identity to get to that fact when it was so glaringly obvious just by looking at the diagram! $\endgroup$ – G-man Nov 7 '15 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.