An acute triangle $ABC$ is given, $AC \not= BC$. The altitudes drawn from $A$ and $B$ meet at $H$ and intersect the external bisector of the angle $C$ at $Y$ and $X$ respectively. The external bisector of the angle $AHB$ meets the segments $AX$ and $BY$ at $P$ and $Q$ respectively. If $PX = QY$, prove that $AP + BQ \ge 2CH$.
Problem
Source: Tuymaada 2021 Senior P7
Tags: geometric inequality, inequalities, geometry
Doru2718
28.07.2021 19:07
This problem is actually false I think,because from the conditions you can show that $AC=BC$
adj0109
28.07.2021 19:29
Doru2718 wrote: This problem is actually false I think,because from the conditions you can show that $AC=BC$ No, the problem is not false. I think $AC = BC$ only if $PQYX$ is an isosceles trapezoid, which must mean $PQYX$ is a parallelogram. @below It can be a parallelogram as well.
Doru2718
28.07.2021 19:35
Isn't $PQYX$ always an isosceles trapezoid? You have $m(\angle HYC)=\frac{\hat C}{2}=m(\angle YHQ)$ so $YX||PQ$.
IndoMathXdZ
30.07.2021 19:13
Great problem. Rename $P,Q$ as $D,E$ respectively. Claim 01. $XY \parallel DE$, which implies $XDEY$ is a parallelogram. Proof. We have \[ \angle CYH = 90^{\circ} - \angle YCB = \frac{1}{2} \angle ACB = \angle YHE \]This implies $XY \parallel DE$. Now, note that $XD = YE$. This implies that $XDEY$ is an isosceles trapezoid. We claim that in fact $XDEY$ is a parallelogram. Suppose otherwise, that $XD$ and $YE$ are antiparallel wrt $DE$. This implies that $\angle XDH = \angle YEH$, and since $\angle XHD = \angle YHE$, this gives us \[ \angle AXB = \angle DXH = \angle HYE = \angle AYB \]which means $AXYB$ is cyclic. However, since $XY \parallel DE$, we have \[ 90^{\circ} - \angle ABC = \angle HAB = \angle YAB = \angle YXB = \angle YXH = \angle XHD = \angle YHE = \angle XYH = \angle XYA = \angle XBA = \angle HBA = 90^{\circ} - \angle CAB \]which gives us $\angle ABC = \angle CAB$, which immediately implies $AC = BC$, which is a contradiction. Claim 02. $AD \cdot EB = XD^2$. Proof. Note that $AX \parallel BY$, which implies \[ \frac{AD}{DX} = \frac{AH}{HX} = \frac{YH}{HB} = \frac{YE}{EB} \]This implies $AD \cdot BE = DX \cdot EY$, and since $XD = YE$, then $AD \cdot BE = XD^2$.
Let $O$ be the circumcenter of $\triangle ABC$. Claim 03. $AX \parallel CO$. Proof. In fact, a more general statement is true. Omitting the condition $XD = YE$, we could say that $AX, CO, BY$ always concur -- and since $AX \parallel BY$, this would imply $AX \parallel CO$. To prove this, note that \[ (CA,CB), (CX,CY) \ \text{are isogonal wrt } XY\]Then by the isogonal line lemma, the line joining $C$ with $AX \cap BY$ and $AY \cap BX = H$ are isogonal. However, we know that $CH$ and $CO$ are isogonal wrt $XY$. This implies, $AX \cap BY, C, O$ are collinear, which implies our result. Claim 04. $AD + BE \ge 2 CH$. Proof. Note that because $AX \parallel CO$, then $CH$ and $DX$ are antiparallel wrt $XY$. Since $HD \parallel CX$, this imply $CHDX$ an isosceles trapezoid, which gives us $CH = DX$. Therefore, by AM-GM inequality and Claim 02, we have \[ AD + BE \ge 2 \sqrt{AD \cdot BE} = 2 \sqrt{DX^2} = 2 \cdot DX = 2 \cdot CH \]
Both Claim 01 and Claim 02 are pretty natural just by angle chasing and length chasing, trying to use every fact from the problem. The core of the problem is in Claim 03. Here's how I think about it. We are asked to prove a geometric inequality $AD + BE \ge 2CH$, and we have gotten the fact that $AD \cdot BE = XD^2$, therefore AM-GM would suggest proving something along the line of $AD + BE \ge 2 \sqrt{AD \cdot BE} \ge 2CH$, which means we should be finish if we have $XD \ge CH$. In fact, proving $XD = CH$ seems a lot better because there's not a lot of geometric inequalities notion that we could derive from $XD$ and $CH$ rather than size of angles, which is not really natural in this sense. So, we are motivated to somehow prove $CH = XD$, i.e. $XD$ and $CH$ are antiparallel wrt $XY$. Indeed, they shouldn't be parallel by diagram. Now, introduce a new point $O$, the circumcenter. Since $CO,CH$ are isogonal wrt $XY$, we should be done if we proved that $AX, CO, BY$ are parallel.
Actually, my proof for $AX,CO,BY$ being parallel is a little bit ugly -- but after talking with my friend, I just realized that isogonal conjugates are pretty motivated in here: there are a lot of isogonal conjugates wrt line $XY$, $(CH,CO),(CA,CB),(CX,CY)$ so we could somehow use isogonal line lemma to force concurrency. This holds for general cases, and therefore we are done.
Alan_Yu_2021
03.08.2021 11:43
I have almost same way as IndoMathXdZ do, but i don't use isogonal line lemma to prove CO//AX to get the CHDX an isosceles trapezoid, i just prove YC/CX=DH/HE.
Mahdi_Mashayekhi
04.08.2024 19:44
It's well-known than angle bisector of $\angle BCA$ and $\angle BHA$ are parallel so $XY \parallel PQ$ and since $PX = QY$ either $PQYX$ is parallelogram or isosceles trapezoid. If $PQYX$ is isosceles trapezoid then $\angle BYX = \angle YXA$ and since $\angle HYX = \angle HXY = \angle \frac{C}{2}$ then $\angle BYA = \angle BXA$ and then $BXYA$ would be cyclic but since $\angle XBA \neq YAB$, we would have contradiction. Since $AX \parallel BY$ we have that $\frac{XH}{HA}=\frac{XB}{YA}$. Note that $\angle CYA = \angle CXB$ and $\angle CAY = \angle CBX$ so $CYA$ an $CXB$ are similar so $\frac{XB}{YA}=\frac{BC}{AC}$. Since $\frac{XH}{HA}=\frac{BC}{AC}$ and $\angle XHA = \angle ACB$ then $AHX$ and $ACB$ are similar so $\angle XAH = \angle A$. Now note that $\angle XPH = \angle A + \angle \frac{C}{2} = \angle CHX + \angle XHP = \angle CHP$ so $CH = XP = YQ$. Since $\frac{XP}{PA} = \frac{BQ}{QY}$ then $\frac{PA}{XP} + \frac{BQ}{QY} \ge 2$ so $\frac{PA}{CH} + \frac{BQ}{CH} \ge 2$ as wanted.