$ABC$ triangle with $|AB|<|BC|<|CA|$ has the incenter $I$. The orthocenters of triangles $IBC, IAC$ and $IAB$ are $H_A, H_A$ and $H_A$. $H_BH_C$ intersect $BC$ at $K_A$ and perpendicular line from $I$ to $H_BH_B$ intersect $BC$ at $L_A$. $K_B, L_B, K_C, L_C$ are defined similarly. Prove that $$|K_AL_A|=|K_BL_B|+|K_CL_C|$$
Problem
Source: Turkey TST 2022 P8 Day 3
Tags: geometry, geometry proposed, orthocenter, incircle
799786
13.03.2022 14:54
How many problems are in a Turkey TST?
electrovector
13.03.2022 14:55
There are 3 days each containing exactly 3 problems which gives a total of 9 problems.
electrovector
13.03.2022 14:56
I will prove that $K_A$ is the incircle touch point and $L_A$ is the midpoint of $BC$. This will clearly finish the problem. Let $D$ be the intouch point and $M$ be the midpoint of $BC$.
Easy angle chase gives $H_CB //H_BC$. Then we have $BH_C: CH_B=BK_A: CK_A$. Let $H_BI \cap AC= E$ and $H_CI \cap AB=F$. We have $BH_C=\frac{BF}{sin \frac{\angle A}{2}}$ and $CH_B=\frac{CE}{sin \frac{\angle A}{2}} \Rightarrow BK_A:CK_A=BD:CD$ which means $K_A=D$.
Now we will prove that $MI \perp H_BH_C$. It is enough to prove that $MH_C^2+ID^2=IH_C^2+DM^2$.
Write $\angle A= 2 \alpha, \angle B=2\beta, \angle C=2\theta$. Using cosines theorem in triangle $BH_CM$, we have $MH_C^2=BH_C^2+BM^2-2 \cdot BM \cdot BH_C \text{cos} (\beta-\theta) \Rightarrow $ we need to prove that
$$BH_C^2+BM^2-BC \cdot BH_C \text{cos}(\beta-\theta)+ID^2=IH_C^2+BM^2+BD^2-BC \cdot BD \overbrace{\Rightarrow}^{\text{sines theorem on} BIH_C}$$$$BI^2 \cdot \frac{\sin^2( \alpha+\theta)}{\sin^2 \alpha}-BC \cdot BI \cdot \frac{\sin(\alpha+\theta)\cos(\beta-\theta)}{\sin \alpha} +BI^2 \cdot \sin^2(\beta)=BI^2 \cdot \frac{\sin^2(\theta)}{\sin^2(\alpha)}+BI^2\cdot \cos^2(\beta)-BC \cdot BI \cdot \cos(\beta) \overbrace{ \Rightarrow}^{\text{sines theorem on BIC and} \alpha+\beta+\theta=90}$$$$\frac{\sin^2(\theta)\cos^2(\beta)}{\sin^2(\beta+\theta)\cos^2(\beta+\theta)}-\frac{\sin(\theta)\cos(\beta)\cos(\beta-\theta)}{\sin(\beta+\theta)\cos(\beta+\theta)}+\frac{\sin^2(\beta)\sin^2(\theta)}{\sin^2(\beta+\theta)}=\frac{\sin^4(\theta)}{\sin^2(\beta+\theta)\cos^2(\beta+\theta)}+\frac{\sin^2(\theta)\cos^2(\beta)}{\sin^2(\beta+\theta)}-\frac{\sin(\theta)\cdot \cos(\beta)}{\sin(\beta+\theta)} \Rightarrow$$$$\sin^2(\theta)\cos^2(\beta)-\sin(\theta)\cos(\beta)\cos(\beta-
\theta)\sin(\beta+\theta)\cos(\beta+\theta)+\sin^2(\beta)\sin^2(\theta)\cos^2(\beta+\theta)=\sin^4(\theta)+\sin^2(\theta)\cos^2(\beta)\cos^2(\beta+\theta)-\sin(\theta)\cos(\beta)\sin(\beta+\theta)\cos^2(\beta+\theta)$$Now writing $\sin(\beta)=x$ and $\sin(\theta)=y$ we need to prove that
$$y^2(1-x^2)-y\sqrt{1-x^2}(\sqrt{(1-x^2)(1-y^2)}+xy)(\sqrt{(1-x^2)(1-y^2)}-xy)(x\sqrt{1-y^2}+y\sqrt{1-x^2})+x^2y^2(\sqrt{(1-x^2)(1-y^2)}-xy)^2=y^4+y^2(1-x^2)(\sqrt{(1-x^2)(1-y^2)}-xy)^2-y \sqrt{1-x^2}(x \sqrt{1-y^2}+y \sqrt{1-x^2})(\sqrt{(1-x^2)(1-y^2)}-xy)^2$$which is obviously true by expanding.
Note: I couldn't write cosines theorem correctly while I was in the exam and I ended something like $2y^2 \sqrt{1-y^2}=0$ which is obviously wrong. Although I realized my mistake , I had only 10 mins left so I couldn't complete my solution:( It is sad.
Attachments:
mrdriller
13.03.2022 16:53
Lemma 1 $H_{B},D,H_{C}$ are collinear. Proof 1 Since $BH_{C}$ and $CH_{B}$ are parallel and $\frac{BD}{DC}=\frac{BF}{CE}=\frac{BH_{C}}{CH_{B}}$ $H_{B},D,H_{C}$ are collinear as desired. Lemma 2 Let M be the midpoint of $BC$. Then $IM\perp H_{B}H_{C}$ Proof 2 Let $AI\cap BH_{C}=X$ and $AI\cap CH_{B}=Y$. It is easy to see that $IXDB$ are cyclic. Let $AI\cap (ABC)=N$. It is known that $NI=NB=NC$ and $BXMN$ are cyclic. Then $\angle{DBX}=\angle{MNI}$. Furthermore since the triangles $BFH_{C}$ and $NMC$, $\frac{BD}{BH_{C}}=\frac{BF}{BH_{C}}=\frac{NM}{NC}=\frac{NM}{NI}$ Thus $MIN \sim DH_{C}B \implies \angle{NIM}=\angle{BH_{C}D}$. Let $IM\cap H_{B}H_{C}=P$. Then $IXPH_{C}$ is cyclic $\implies$ $IP \perp H_{B}H_{C}$. From this, we can immediately see $K_{A}=D$ and $L_{A}=M_{A}$ where $M_{A}$ is the midpoint of $BC$. Similar for other points and the conclusion is obvious.
GuvercinciHoca
13.03.2022 18:01
Here is an amazing approach to this problem by serdarbozdag, hakN, sevket12, cookierockie, bariskoyuncu, infinityfun: Let $M_A,M_B,M_C$ be the midpoints of $BC,AC,AB$ respectively and incircle of $\triangle ABC$ touches $BC,AC,AB$ at $D,E,F$ respectively. Claim: $H_BH_C$ is the polar of $M_A$ wrt incircle. Proof: Let $EF\cap M_AM_C=N.$ Then by iran lemma we know $C,I,N$ are collinear and $IC\perp AH_B.$ Also we know that $N$ lies on the polar of $A$ wrt incircle. Thus, $AH_B$ is the polar of $N$ wrt incircle. By La-Hire's Theorem, $N$ lies on the polar of $H_B.$ Which implies $M_CM_A$ is the polar of $H_B$ because $H_BI\perp M_CM_A.$ Similarly we can get $M_BM_A$ is the polar of $H_C.$ So again by La-Hire's Theorem, $H_BH_C$ is the polar of $M_A$ wrt incircle. After this claim we can easily say $L_A=M_A$ and $K_A=D$ (because $M_AD$ is tangent to incircle.) And the rest follows.
SerdarBozdag
13.03.2022 18:18
For generalization and more properties see https://artofproblemsolving.com/community/c6h1252050p6455626.