Let $ ABC$ be a triangle with $ \angle A < 60^\circ$. Let $ X$ and $ Y$ be the points on the sides $ AB$ and $ AC$, respectively, such that $ CA + AX = CB + BX$ and $ BA + AY = BC + CY$ . Let $ P$ be the point in the plane such that the lines $ PX$ and $ PY$ are perpendicular to $ AB$ and $ AC$, respectively. Prove that $ \angle BPC < 120^\circ$.
Problem
Source: APMO 2008 problem 1
Tags: geometry, Euler, circumcircle, incenter, geometric transformation, reflection, trigonometry
Nemion
22.03.2008 19:46
Consider the triangle of the excenters $ O_1O_2O_3$ $ P$ is its circumcenter. $ ABC$ is the orthic triangle of $ O_1O_2O_3$. Then, $ P$ lies in the Euler line of $ O_1O_2O_3$ in fact the circumcenter $ O$ of the orthic triangle, $ ABC$ it's halfway from the orthocenter of $ O_1O_2O_3$ ($ I$) and $ P$. By easy angle computations and using the fact that $ <A<60$° is easy to see that $ <BIC<<BOC$ then, because $ P$ is located in the same line as $ I$ and $ O$ (in direction to $ O$ from $ I$) its angle must be even less than $ <BOC=2<A<120$°
andersonw
16.03.2010 00:40
Sorry for bumping, but I can't see why P is the circumcenter of the excentral triangle. Could someone explain why to me? Also, shouldn't P be located in the direction from I to O, not the other way, because the Euler line has the nine point center (circumcenter of ABC) halfway between the circumcenter (point P) and orthocenter (incenter of ABC)? Or am I just being stupid?
SnowEverywhere
21.03.2010 19:31
Does this solution work? Or do I have to prove some of the stuff in paragraph 2 as Lemmas? Let $ I$ be the incenter of triangle $ ABC$ and let the incircle of $ ABC$ touch $ AB$ and $ AC$ at $ X'$ and $ Y'$ respectively. Let $ M$ and $ N$ denote the midpoints of $ AB$ and $ AC$ and $ O$ the circumcenter of $ ABC$. Note that by definition, $ BC + AX' = AC + BX'$ and therefore it follows that $ AX' = BX$. Therefore $ X$ is the reflection of $ X'$ in $ M$. Similarly, $ Y$ is the reflection of $ Y'$ in $ N$. Hence $ PX$ is the reflection of $ IX'$ in $ OM$ and $ PY$ is the reflection of $ IY'$ in $ ON$. Therefore $ P$ is the reflection of $ I$ in $ O$. Now consider the circumcircle $ \Gamma$ of $ OBC$. Since $ \angle{A} < 60$, it follows that $ \angle{BOC}= 2\angle{A}<90+\angle{A}/2 = \angle{BIC}$. Hence point $ I$ is in the interior of $ \Gamma$. Therefore $ P$, the reflection of $ I$ in point $ O$ which lies on $ \Gamma$, must be in the exterior of $ \Gamma$. Therefore it follows that since $ P$ is in the exterior of $ \Gamma$, that $ \angle{BPC}<\angle{BOC}=2\angle{A}<120$.
utkarshgupta
24.01.2015 13:47
Call the excentres of the $\triangle ABC$ $I_A,I_B,I_C$ Then quite obviously, $P$ is the circumcentre of $\triangle I_AI_BI_C$ Call $I_BB \cap I_CC = H$ Then again obviously $H$ is the orthocentre of triangle $I_AI_BI_C$ Claim : $\angle BPC \le \angle BHC$ If we prove $P$ lies outside $\odot BHC$, we are done. We know $I_ABHC$ is cyclic with $I_AH$ as diameter. But $I_AH = 2I_AP \cos(90-\frac{A}{2})=2I_AP \sin (\frac{A}{2}) < I_P$ $\implies P$ lies outside $\odot BHC$ $\implies \angle BPC < \angle BHC = 90+\frac{A}{2} < 120$ Thus we are done
MathPanda1
30.01.2016 04:51
What is the motivation for looking at $I$ and $O$ and the circumcircle of $BOC$? Thanks!
anantmudgal09
06.02.2017 17:00
Nice! Let $I$ be the incenter of $\triangle ABC$. Note that $P$ is the circumcenter of the excentral triangle $\triangle J_AJ_BJ_C$ of $\triangle ABC$. Secondly, $$J_AI=2J_AP \cos \left(90^{\circ}-\frac{A}{2}\right)=2J_AP \sin \left(\frac{A}{2}\right)>J_AP,$$so $P$ lies outside the circumcircle of $\triangle BIC$. Consequently, $\angle BPC<\angle BIC=\left(90^{\circ}+\frac{A}{2}\right)<120^{\circ}$ as desired.
vsathiam
09.04.2017 21:48
MathPanda1 wrote: What is the motivation for looking at $I$ and $O$ and the circumcircle of $BOC$? Thanks! The length condition suggests excenters.
Shaddoll
30.04.2017 05:03
Note that $X$ is the contact point of the $C$-excircle and $Y$ is the contact point of the $B$-excircle(since we get $AX = s-b$ and $AY = s-c$). Thus, it suffices to consider the excentral triangle $I_AI_BI_C$. Let $K$ be the point on $(I_AI_BI_C)$ such that $KI_BI_C$ is isosceles.
Lemma 1: $P$ is the circumcenter of $I_AI_BI_C$
Proof: $\angle PI_BI_C = 90-\angle I_BAC = 90-\angle I_BI_AI_C$, and similarly, $\angle PI_CI_B = 90-\angle I_BI_AI_C$, and we know that the circumcenter satisfies the $2$ angle conditions, and only $1$ point can simultaneously satisfy the $2$ angle conditions(since it is the intersection of $2$ lines).
Now, let $H_K$ denote the orthocenter of $KI_BI_C$. Let $H$ be the orthocenter of $I_AI_BI_C$.
Lemma 2: $KH_K = I_AH$
Proof: We use complex numbers, letting $(I_AI_BI_C)$ as the unit circle. Note that $h_K = k+b+c \implies h_K-k = b+c$. Furthermore, $h = a+b+c \implies h-a = b+c$.
Lemma 3: $\angle I_BI_AI_C > 60^{\circ}$
Proof: Note that $\angle I_BAC = \angle I_BI_AI_C$ and similarly, $\angle I_CAB = \angle I_BI_AI_C$, so $\angle BAC = 180-2\angle I_BI_AI_C$. Since $\angle BAC < 60^{\circ}$, this implies $\angle I_BI_AI_C > 60^{\circ}$.
Furthermore, we have $\triangle I_AI_BI_C$ is acute, since $\angle BAC = 180-2\angle I_BI_AI_C$, implying $\angle I_BI_AI_C$ is acute, and analogously for the other angles. Now, we claim that $P$ lies outside $(BCH)$. Note that since $\angle I_ABH = \angle I_ACH = 90^{\circ}$, this is just the circle with diameter $I_AH$.
Lemma 4: $P$ lies outside of the circle with diameter $KH_K$
Proof: Note that $\angle I_BPI_C = 2 \angle I_BI_AI_C$. Furthermore, $\angle I_BH_KI_C = 180-\angle I_BI_AI_C$ by cyclic quadrilaterals, so since $\angle I_BI_AI_C > 60^{\circ}$, this means that $\angle I_BPI_C > \angle I_BH_KI_C$, and thus $P$ lies closer to $BC$ than $H$, and thus cannot lie inside $(BCH)$.
Now, note that this means that $R > KH_K = I_AH$, where $R$ is the circumradius of $I_AI_BI_C$. Thus, $P$ lies outside of the circle with diameter $I_AH$, since $I_AP = R$. However, the locus of points $N$ where $\angle BNC > 120^{\circ}$ is clearly contained within the circle diameter $I_AH$, since $\angle BMC$, where $M$ is on the circle with diameter $I_AH$, is at most $120^{\circ}$, as it is either equal to $BI_AC$ or $BHC = 180-BI_AC$. Thus, we must have $\angle BPC < 120^{\circ}$.
Why-
11.02.2020 16:22
You can find that $O$ is the midpoint of $I$ and $P$ So it can be solved easily.
im.a.robot
23.11.2020 21:36
P= the circumcenter of IaIbIc O circumcenter of ABC =the center of 9point circle of IaIbIc I the incenter of triangle ABC = the orthocenter of triangle IaIbIc So PI is th euler line of IaIbIc so P,O,I are on the same line. So point p is out of triangle IBC so we have that the angle BPC≤the angle BIC=90+A/2≤120 and the problem is easelly solved