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Construction

Here the construction of the diagram.

  1. Draw a circle $C_1$ as incircle of triangle $ABC$. $I$ is the incenter and H is orthocenter of triangle $ABC$.
  2. Draw circle $C_2$ as circumcircle of triangle $BIC$. Here $C_2$ intersects $AC$ at $D$ and extension of $AB$ at $E$.
  3. Draw circle $C_3$ as circumcircle of triangle $ADE$.
  4. Draw circle $C_4$ as circumcircle of triangle $ABC$. Circle $C_3$ and circle $C_4$ intersect each other at $P$.

Given that $AB=10, AC=16$ and $CE=16$ what is the length of $HP$?

First of all, I apologize if my diagram drawn badly. I spent hours to draw and redraw to give better perspective. Whenever I draw I always obtain $C_4$ and $C_3$ pretty much same in size. It made me wonder if the size would be the same in every condition. But I cannot figure out how to prove that claim. If it is proven, $A,I$ and $P$ must be collinear right?

Secondly, I have initiative to make circumcenter of triangle $ABC$ and label it as $J$ considering that $H,I$ and $J$ must be colinear and maybe I could make use property of orthocenter and incircle namely $IJ=2HI$. So here I draw triangle $HJP$ hoping it would help to solve the problem. I also have a claim that triangle $HJP$ somehow similar to triangle $ABC$. But again, I cannot prove my claim.

I also found that triangle $ABC$ cannot be right-angled triangle since it make $CE$ can never be 16. (I tried choose special case to make it easier in calculation since there is no restriction about the type of the triangle).

I cannot figure out any further ideas. I very appreciate for every responses, ideas and answer you share. Thankyou in advance.

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    $\begingroup$ A triangle's incenter, orthocenter, and circumcenter are not necessarily collinear. However, the centroid, orthocenter, and circumcenter lie on the Euler line. $\endgroup$ Commented Nov 19, 2024 at 16:50
  • $\begingroup$ Oh, really? Is not centroid and Incenter supposed to be the exactly same point in every triangle?@Blue $\endgroup$ Commented Nov 19, 2024 at 22:41
  • $\begingroup$ I had done my research, and yes you are correct, centroid isn’t necessary the same point with Incenter. Thanks for let me know. @Blue $\endgroup$ Commented Nov 19, 2024 at 23:01
  • $\begingroup$ "Is not centroid and Incenter supposed to be the exactly same point in every triangle?" They're distinct, except in equilateral triangles. FYI: The incenter is $X(1)$ in the Encyclopedia of Triangle Centers; the centroid is $X(2)$. (The circumcenter and orthocenter are $X(3)$ and $X(4)$.) $\endgroup$ Commented Nov 19, 2024 at 23:08

2 Answers 2

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Let's show, first of all, that triangle $ACE$ is equilateral. As $AC=CE$ we also have $\angle CAE=\angle CEA=\alpha$. On the other hand: $$ \angle CIB=180°-{\angle ACB\over2}-{\angle CBA\over2}=90°+{\alpha\over2}. $$ But $\angle CIB$ and $\angle BEC$ both insist on the same chord $CB$ of circle $BIC$, on opposite arcs. Hence $\angle CIB = 180°-\angle BEC$, that is: $$ 90°+{\alpha\over2}=180°-\alpha, \quad\text{which gives:}\quad \alpha=60°. $$ It follows that $ACE$ is equilateral, as it was to be proved.

Note that, by symmetry, $AD=AB=10$. Moreover, orthocenter $H$ also lies on circle $BIC$, because (see figure below):

$$ \angle CHB=\angle GHF=180°-\alpha=120°. $$

Circle $ADE$ is congruent to circle $BIC$. To see why, reflect triangle $AEC$ about $EC$ and let $J$, $K$ the intersections of the reflected triangle with circle $BIC$. It is not difficult to prove that $BC=CK=KB$ (all these chords are subtended by $120°$ inscribed angles) and thus $CJ=BD=AD$. It follows that trapezoid $ADLE$ is the same as trapezoid $CJKE$, rotated by $60°$ about point $E$.

Circle $ADE$ is then the image of circle $BIC$ under a rotation of $60°$ about point $E$. This implies that the center of circle $BIC$ lies on circle $ADE$. Analogously, circle $ABC$ is the reflection of circle $ADE$ about line $AI$ (the bisector of $\angle EAC$), hence the center of circle $BIC$ also lies on circle $ABC$, and is thus point $P$.

$PH$ is then the radius of circle $BIC$. To compute it, we can find $DE=14$ by the cosine rule applied to triangle $ADE$ and then: $$ PH={DE\over2\sin\alpha}={14\over\sqrt3}. $$

enter image description here

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    $\begingroup$ What a well explained answer. Thank you very much $\endgroup$ Commented Nov 20, 2024 at 2:23
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enter image description here

Particular case, where triangle ABE is equilateral. In this case as can be seen in figure circle $C_2$ passes vertexes A and C and the orthocenter of triangle ABC and its radius is equal to that of circumcircle of triangle ABC and $PH=R$. So we have to find R.We have:

$\angle ABC=60^o$

$\angle ACE_1=30^o$

therefore:

$AE_1=\frac{16}2=8\Rightarrow BE_1=10-8=2$

$CE_1=\frac{\sqrt3}2\times 16=8\sqrt 3$

$CB=\sqrt{(8\sqrt 3)^2+2^2}=14$

Area of triangle ABC is:

$S=\frac{(CE_1=8\sqrt 3)\times 10}2=40\sqrt 3$

Finally R is:

$R=\frac{abc}{4s}=\frac{16\times 14\times 10}{4\times 40\sqrt 3}=\frac{14\sqrt 3}3$

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  • $\begingroup$ I should have known to work on equilateral triangle for the particular case rather than right-angled triangle. Anyway thankyou for the answer. @sirous $\endgroup$ Commented Nov 19, 2024 at 22:43
  • $\begingroup$ In addition, a very good diagram you attach really made a very clear understanding about the problem. But, I wonder is there a mathematical reason why in this case PH must be equal to R, except only deduced from the diagram? @sirous $\endgroup$ Commented Nov 19, 2024 at 22:59

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