Oh man, a geometric functional equation - interesting.

djmathman wrote: Oh man, a geometric functional equation - interesting.

You need to show that a 45-45-90 triangle is great as well (which is not completely trivial).

Oops that is true. It seems that I have forgotten the first fundamental rule of functional equations - one must always plug the solution back in to make sure it checks!

djmathman wrote: With this in mind, set $D$ to be the foot of the altitude from $A$ to $\overline{BC}$... Just to put it out there, it is fairly easy to length-chase to get $AB=AC$ from here by using polynomials.

You can actually take D to be midpoint of BC, then to prove AB=AC will be easier, because then the feet of D to AB and AC is the midline,and D'A will then be parallel to BC... (of course, this is after you proved BAC=90 using angle bisector.)

First of all choose $D$ to be the foot of angle bisector from $A$. From here we obtain $\angle BAC=90$ by angle chasing. Now use the first observation and choose $D$ to be the foot of perpendicular from $A$. Now by length chasing and triangle similarity, we obtain $AB=AC$.

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First we select $D$ as the intersection of $AO$ and $\overline{BC}.$ Let $E$ denotes the foot from $A$ to $\overline{BC},$ and $D^*,E^*$ denote the reflection of $D,E$ in line $PQ$ respectively. Since $PQ$ is parallel to $BC,$ it's easy to see that $E^*$ is the orthocenter of $\triangle APQ,$ which implies that $E^*$ and $D^*$ are symmetry over the perpendicular bisector of $\overline{BC},$ in other words, \[D^*\text{ lies on the circumcircle of }\triangle ABC\iff\angle BD^*C=\angle BE^*C=\angle A\iff E^*\equiv A.\]Therefore $PQ$ is median line $\implies \angle A=90^\circ.$ This is because $D$ is the intersection of the perpendicular bisector of $\overline{AB}$ and $\overline{AC}.$ [asy][asy] size(7cm); pointpen=black; pathpen=black; pointfontpen=fontsize(9pt); void b(){ pair A=D("A",dir(110),dir(110)); pair B=D("B",dir(-150),dir(-150)); pair C=D("C",dir(-30),dir(-30)); pair D=D("D",extension(A,origin,B,C),S); pair P=D("P",foot(D,A,B),W); pair Q=D("Q",foot(D,A,C),E); pair E=D("E",foot(A,B,C),S); pair D1=D("D^*",2*foot(D,P,Q)-D,dir(120)); pair E1=D("E^*",2*foot(E,P,Q)-E,N); D(A--B--C--cycle); D(P--D--Q,magenta); D(B--E1--C); D(B--D1--C); D(P--Q,dashed); draw(anglemark(B,E1,C,3),deepgreen); draw(anglemark(B,D1,C,3),deepgreen); D(E1--E,dotted); D(D1--D,dotted); D(circumcircle(A,P,Q),red+linetype("4 4")); } b(); pathflag=false; b(); [/asy][/asy] Then we select $D$ as the foot from $A$ to $\overline{BC}.$ By the above result we get $APDQ$ is a rectangle, thus $D^*\in \odot(APDQ),$ and \[\begin{aligned}D^* \text{ lies on the circumcircle of }\triangle ABC &\iff D^* \text{ sends $\triangle D^*PB$ to $\triangle D^*QC$}\\ &\iff \frac{DP}{PB}=\frac{D^*P}{PB}=\frac{D^*Q}{QC}=\frac{DQ}{QC}\\ &\iff \tan{\angle B}=\tan{\angle C}\\ &\iff \angle B=\angle C. \end{aligned}\]as desired. $\square$ [asy][asy] size(7cm); pointpen=black; pathpen=black; pointfontpen=fontsize(9pt); void b(){ pair A=D("A",dir(70),dir(70)); pair B=D("B",dir(180),W); pair C=D("C",dir(0),E); pair D=D("D",foot(A,B,C),S); pair P=D("P",foot(D,A,B),W); pair Q=D("Q",foot(D,A,C),S); pair D1=D("D^*",2*foot(D,P,Q)-D,dir(70)); D(P--D--Q,magenta); D(D--D1,dotted); D(B--P--D1--cycle,blue+dashed); D(D1--Q--C--cycle,blue+dashed); D(A--P--Q--cycle); D(B--C); } b(); pathflag=false; b(); [/asy][/asy]

As we have already seen we get $\angle BAC=90^{\circ}$ by setting $D$ to be the foot of angle bisector of $\angle BAC$. Now let $D$ be the midpoint of $BC$, then $P,Q$ are the midpoints of $AB,AC$ respectively as $\angle BAC=90^{\circ}$. Now let $D'$ be the reflection of $D$ in $PQ$. Then $\angle PD'Q=\angle PDQ=\angle PAQ\implies AD'PQ$ is cyclic, thus $D'\equiv\odot(ABC)\cap\odot(APQ)$. But $\odot(APQ)$ is tangent to $\odot(ABC)$ at $A$, hence $D'\equiv A$, hence $AD\perp PQ\implies AB=AC$, as required.

Let $E$ be reflection of $D$ in $PQ$. 1. Suppose $\triangle ABC$ is great: Let $D$ be intersection of bisector of $\angle A$ and $BC$. $P,Q$ are symmetric wrt. $AD$, so reflection of $D$ in $PQ$ lies on $AD$. Both $A$ and $E$ lie on $\odot ABC$ and they are on same side of $PQ$, hence $A\equiv E \Longrightarrow \angle PAQ=\angle PDQ$, but since $APDQ$ is cyclic, $\angle PAQ=180^{\circ}-\angle PDQ$, so $\angle BAC=\angle PAQ=90^{\circ}$. Now let $D$ be midpoint of $BC$. Let line paralel to $BC$ through $A$ intersect $\odot ABC$ at $A'$. Since $APDQ$ is a rectangle, we have $d(A,PQ)=d(D,PQ)=d(E,PQ)$, so $AE\parallel BC\Longrightarrow E\equiv A'$.Let $K,L$ be projections of $A,A'$ on $BC$, $AA'BC$ is isosceles trapezoid, so $K,L$ are symmetric wrt. $D$. Then since $K\equiv D\Longrightarrow L\equiv D\Longrightarrow A=A'$ so $\triangle ABC$ is $A$-isosceles and $\angle A=90^{\circ}$. 2. Suppose $\angle A=90$ and $AB=AC$. $\angle PEQ=\angle PDQ=90^{\circ} \Longrightarrow PEDQ$ is cyclic. $QE=QD=AP=QC$, and $PE=PD=AQ=PB$. $\angle EQC=\angle EQD+90^{\circ}=180^{\circ}-\angle EPD+90^{\circ}=\angle BPE$ $\Longrightarrow \triangle QEC\sim \triangle PEB $ $\Longrightarrow \angle BEC=\angle PEQ=90^{\circ}\Longrightarrow E\in \odot ABC$

Sorry! I read all post on the top carefully But I have a question with case "ABC is great $\rightarrow$ $\angle A=90^{\circ}$ and $AB=AC$" If D be a random point on the side BC (is not midpoint BC, foot of the angle bisector of $\angle A$,..) then prove that $\angle A=90^{\circ}$ and $AB=AC$????? Please answer this my question! Thanks all!

^^ That doesn't matter. We've shown that by setting $D$ to be the foot of the angle bisector or the midpoint of $BC$ or the foot of the altitude or whatever that the only triangle $\triangle ABC$ can possibly be is an isosceles right triangle. Thus, we don't have to test all the other points to conclude this. Let's go off on what seemingly might be a tangent. Consider the following functional equation: find all functions $f$ such that \[f(x)+f(y)=x^2+y^2\]for all real $x$ and $y$. This FE has a very simple solution: substitute $y=x$ to obtain \[2f(x) = 2x^2\quad\implies\quad f(x) = x^2.\]Now we just need to check that $f(x)$ always works, which it does. This may seem like a strange example, but the point is made clear: in order to show that $f(x)=x^2$ is the only possible function which satisfies the given conditions, all we need to do is to look at the case where $x=y$. Once we have that, we just need to check that $f(x)=x^2$ works. The same thing goes with this problem. In all the posts above, we use the special cases of $D$ being the angle bisector/midpoint/etc. to deduce that $\triangle ABC$ must be an isosceles right triangle. Once we have that, we just need to check that indeed $\triangle ABC$ is great - which I do in post 4.

Dear djmathman, Thanks you very much! I understand this problem is a geometric functional equation! But, Sorry, I still a small question. I think we must see only case such as midpoint or angle bisrctor or... No need see all case.

First, we prove the if direction. Denote the reflection of $D$ wrt $PQ$ be $D'$. Clearly $D'AQP$ is cyclic, so $\angle D'PB = \angle D'QC$. Using $D'P =PD = BP$ and $D'Q=QD=QC$ gives $\triangle D'PB \sim \triangle D'QC$, so $\angle BD'C = \angle PD'Q=90$. This implies that $AD'BC$ is cyclic. We are done. Now we prove the only if direction. First, take $D$ where $AD$ bisects $\angle A$. Then $\triangle PAD \equiv \triangle QAD$, so $D'$ lies on $AD$. So if $AD'CB$ cyclic gives $A=D'$, so $AQDP$ is a square, so $\angle A = 90$. Take $D$ be the perpendicular from $A$ to $BC$. $D'$ being on $(ABC)$ is equivalent to $\triangle D'PB \sim \triangle D'QC$. This is equivalent to $\frac{DP}{PB} = \frac{DQ}{DC}$, or $\angle B = \angle C$. We are done.

Let $D'$ denote the reflection of $D$ in $PQ$. First, take the foot of the angle bisector from $A$. Then we see that $\angle A = 90^\circ$ as $A$ and $D'$ lie on the same side of $BC$. Now, we know $\angle PDQ = 90^\circ$, so it suffices to show $\angle BD'P = \angle CD'Q$. However we already know that $\angle D'PB = \angle D'QC$ as $AD'PQD$ is cyclic, so this is equivalent to the condition \[ \frac{D'P}{PB} = \frac{DP}{PB} = \frac{DQ}{QC} = \frac{D'Q}{QC}\]However, we have \[ \frac{DP}{PB} = \frac{QC}{DQ}\]since $PD \parallel AC$, so this is equivalent with $BPD$ and $CQD$ being isosceles right. But they are similar to the original, so the original condition is equivalent with $ABC$ being isosceles right with $\angle A = 90^\circ$.

Can we use continuity to argue that A must be the image of some point on BC? As B is its own image, and so is C, and we can argue that the image of a point on BC that is very close to B must lie on arc BAC, very close to B, so as point D travels from B to C on BC, its image travels from B to C on arc BAC, and therefore its image must overlap with A at some point. Is this valid?

shinichiman wrote: We say that a triangle $ABC$ is great if the following holds: for any point $D$ on the side $BC$, if $P$ and $Q$ are the feet of the perpendiculars from $D$ to the lines $AB$ and $AC$, respectively, then the reflection of $D$ in the line $PQ$ lies on the circumcircle of the triangle $ABC$. Prove that triangle $ABC$ is great if and only if $\angle A = 90^{\circ}$ and $AB = AC$. Senior Problems Committee of the Australian Mathematical Olympiad Committee Anyone tried coordinate bash? Taking $A(0,0),B(0,y_1)$ and $C(x_2,y_2)$ simplifies the problem quite well, and it is easier to interpret reflection in tje Cartesian plane. I will post the solution as soon as I find one.

shinichiman wrote: We say that a triangle $ABC$ is great if the following holds: for any point $D$ on the side $BC$, if $P$ and $Q$ are the feet of the perpendiculars from $D$ to the lines $AB$ and $AC$, respectively, then the reflection of $D$ in the line $PQ$ lies on the circumcircle of the triangle $ABC$. Prove that triangle $ABC$ is great if and only if $\angle A = 90^{\circ}$ and $AB = AC$. Senior Problems Committee of the Australian Mathematical Olympiad Committee Easy problem for APMO. Let $D = BC \cap$ Angle Bisector of $\angle BAC$. It can be also seen that $D_1$ lies on $AD$. Since $A, D$ lie on opposite sides if $PQ$, $A, D_1$ lie on same side of $PQ$. But these assertion imply that $A = D_1$ which means that $APDQ$ is a square which means that $\angle BAC = 90^\circ$. Consider $D$ on $BC$ such that $\angle ADB = 90^\circ$. Then using the previous condition and this condition we get that $AB=AC$ in two steps because we get that $D = \odot (ABC) \cap \odot (APQ)$. Now to prove this claim works, observe that it is angle chasing after noticing that $P, Q$ are circumcenter of $\triangle BDD_1$ and $\triangle CDD_1$ respectively.

Why is this so hard? I sincerely found it much harder than APMO 16 P2,4,5 Let $D'$ be the reflection of $D$ over $PQ$. If: note $\angle PDB=\angle QDC=45^{\circ}$, so $\angle PDQ=\angle PD'Q=90^{\circ}$ note $BP=PD=PD'$, and $CQ=DQ=D'Q$. We can see that $\angle BD'C=\angle PD'Q-\angle PD'B+\angle QD'C$. Since $BP=PD', D'Q=QC$, $\angle PD'B=\frac 12 \angle APD'$ and $\angle QD'C=\frac 12 \angle AQD'$. Since $\angle BAC=\angle PAQ=\angle PD'Q=90^{\circ}$, $APD'Q$ is cyclic, so $\angle APD'=\angle AQD'$, as desired. Only if: let $D$ be the foot of the $\angle A$'s bisector. Then $\angle PAD=\angle QAD, AD=AD, \angle APD=\angle AQD$, so $APD$ is congruent to $AQD$, so $PD=QD$. This implies $AD$ perpendicularly bisect $PQ$, so $PR=RQ$. Therefore, $D'$ lies on $AD$, so $D'=A$, and $AR=RD$, so $\angle BAC=90^{\circ}$. Furthermore, $APDQ$ is actually a square. To show $AB=AC$, I take $D=O$, the midpoint of $BC$. Then notice $OP,OQ$ are the midlines of $\triangle ABC$, and so is $PQ$. The key observation is $AD'||BC$. Indeed, let $R=AQ\cap D'P$, we can see $AR+RQ=D'R+RP$ and $AR\cdot RQ=D'R\cdot RP$, so it follows that $AR=D'R, QR=PR, AD||PQ||BC$. Furthermore, $D'O\perp BC$, so if $D'$ is on the circumcircle and $BC$ is horizontal, $D'$ is the highest point (or the lowest point) on the circle. Since $AD'||BC$, it follows that $A=D'$.

First we'll prove other triangles are not Great then we'll prove 90-45-45 is Great. we just need to find special points as D. Let AD be angle bisector of ∠A. It's well known AD is perpendicular bisector of PQ so D' must be A. ∠PAQ = ∠PDQ = ∠ACB + ∠ABC so ∠PAQ = 90 so ∠A = 90. Now let AD be altitude. APDQ is rectangle so AD'PQ is isosceles trapezoid. ∠D'AP = ∠APQ = ∠ACB so if AD'BC is cyclic we have ∠AD'B = 180 - ∠ACB = 180 - ∠D'AP so D' lies on AB and is A so AD is perpendicular to APDQ so APDQ is square so AB = AC. Let ABC be a 90-45-45 triangle and D a random point on BC. PB = PD = PD' and QC = QD = QD' so P and Q are center of BD'D and CD'D. ∠BD'C = ∠BD'D + ∠CD'D = 45 + 45 = 90 so BAD'C is cyclic. we're Done.

Claim: $\angle A=90$ and $AB=AC$ is necessary. Proof. For each $D_i$ let $P_i$ and $Q_i$ be the feet from $D_i$ to $\overline{AB}$ and $\overline{AC}.$ Let $D_1$ be the foot of the angle bisector from $A$ and note that $D_1'$ lies on $\overline{AD_1}.$ Hence, $D_1'=A$ and since $AP_1D_1Q_1$ is cyclic, $$2\angle BAC=\angle P_1AQ_1+\angle P_1D_1Q_1=180$$and $\angle A=90.$ Let $D_2$ be the foot from $A$ to $\overline{BC}.$ Then, $D_2'$ lies on $(P_2D_2Q_2)$ and $(ABC)$ so $D_2'=A.$ Hence, $AP_2D_2Q_2$ is a square and $AB=AC.$ $\blacksquare$ Claim: $\angle A=90$ and $AB=AC$ is sufficient. Proof. Notice $D'P=DP=BP$ so $\angle DD'B=\tfrac{1}{2}\angle DPB=45.$ Similarly, $\angle DD'C=45$ so $\angle CD'B=90=\angle A.$ $\blacksquare$ $\square$

First, we assume $ABC$ is great. Let $D_1$ denote the reflection of $D$ in $PQ$. Suppose $D$ is the foot of the internal bisector of $\angle BAC$. Then, we clearly have $DP = DQ$, so $APDQ$ is a kite. It follows that $AD \perp PQ$, so $A$ and $D_1$ must coincide. Thus, $$PA = PD = QD = QA$$which means $APDQ$ is a square, as $AD \perp PQ$. As a result, we know $\angle A = 90^{\circ}$ must hold. Now, suppose $D$ is the midpoint of $BC$. Because $APDQ$ is a rectangle, $$\angle PD_1Q = \angle PDQ = 90^{\circ} = \angle PAQ$$so $APDQD_1$ is cyclic with diameter $PQ$. Moreover, $$\angle APQ = \angle PQD = \angle PQD_1 = \angle PAD_1$$implies $AD_1QP$ is a cyclic isosceles trapezoid. Hence, $AD_1$ and $PQ$ have the same perpendicular bisector, so $DP = DQ$ follows from $DA = R = DD_1$. Thus, $$AB = 2 \cdot DQ = 2 \cdot DP = AC$$which means the only possible solution is $\angle A = 90^{\circ}$ and $AB = AC$. Now, we show that $ABC$ is indeed great when $\angle A = 90^{\circ}$ and $AB = AC$. Let $M$ be the circumcenter of $ABC$. It's clear that $AD_1QP$ is still an isosceles trapezoid. Now, since $BDP$ and $CDQ$ are also isosceles right triangles, we have $$BP \cdot PA = PD \cdot DQ = AQ \cdot QC$$so $$MP^2 = Pow_{(ABC)}(P) + R^2 = Pow_{(ABC)}(Q) + R^2 = MQ^2.$$Thus, $MA = MD_1$ follows from $MP = MQ$, which finishes. $\blacksquare$ Remarks: The flavor of this question is similar to that of ISL 2020/G1. In fact, attempting to solve that problem definitely helped me solve this question.

Send $D$ to $B$ and $C$ respectively, and the reflection $D'$ gets sent to $B$ and $C$ respectively as well. Therefore, by continuity, there exists some choice of $D$ such that $D'$ has the same "x-value" as $A$ if the x-axis is set to $\overline{BC}$ (i.e. $\overline{AD'} \perp \overline{BC}$ or $A=D'$). This clearly means that $D'=A$, since $D'$ is clearly not the second intersection of the $A$-altitude with $(ABC)$. But on the other hand, $\angle PD'Q=\angle PDQ=180^\circ-\angle PAQ$, so we must have $\angle A=90^\circ$. Now, if $\triangle ABC$ is right, let $D$ be the midpoint of $\overline{BC}$, so $P$ and $Q$ are the midpoints of their respective sides as well. Then $d(D',\overline{BC})=2d(D,\overline{PQ})=d(A,\overline{BC})$. Since $D'B=D'C$, this is bad unless $A$ is the arc midpoint of $\overline{BC}$, i.e. $AB=AC$. $\blacksquare$

When $D = B$, we have $D' = B$ and when $D = C$, we have $D' = C$. Thus move $D$ continuously from $B$ to $C$, then since $D'$ lies on the same side of $BC$ as $A$, there exists $E$ on $BC$ so that when $D = E$, we have that $D' = A$. However this implies $A, E$ are equidistant from $P, Q$ when $D = E$, so $PQ$ passes through their midpoint, however $AE$ is a diameter of $(APEQ)$ so it follows that $\angle BAC = \angle PAQ = 90^{\circ}$. Now, set $D$ to be the midpoint of $BC$, then the reflection of $D$ over $PQ$ is the projection of $A$ onto the perpendicular bisector of $BC$, since this must lie on $(ABC)$, it follows that $A$ lies on the horizontal bisector of $BC$, as desired. Now to complete the case, when $ABC$ is as desired, we have that $AD$ passes through the center of $(APQ)$, so it is isogonal in $\angle A$ to the perpendicular to $PQ$. Thus if $A'$ is the reflection of $A$ over $BC$, it follows that $DA', AD$ are isogonal in $\angle A$ so $DA' \perp PQ$. Let $S = DA' \cap (ABC)$, then it follows that $SA \parallel PQ$. Since $PQ$ is a diameter of $(APQD)$, it also follows that $A, D'$ lie on the same side of $PQ$ and $d(A, PQ) = d(D', PQ)$, so we have $AD' \parallel PQ$ as well. Thus $S = D'$, finishing.

If $D$ is arbitrarily close to $B$, then the reflection of $D$ over $PQ$ lies on a line arbitrarily close to the reflection of $BC$ over the $B$-altitude and is arbitrarily close to $B$, which means that the $B$-altitude is the angle bisector of the tangent to $B$ is the angle bisector of $BC$, so $\angle B=\frac12\angle A$. Similarly, $\angle C=\frac12\angle A$, so $ABC$ is an isosceles right triangle. If $ABC$ is an isosceles right triangle, then $APDQ$ is a rectangle so if $O$ is the midpoint of $BC$ and $D'$ is the reflection of $D$ over $PQ$, then $APDQOD'$ is cyclic, so $\angle BAO=\angle OAC$ implies $\angle PD'O=\angle QAO$, and $D'Q=DQ=AP$ implies $\angle AD'P=\angle D'AQ$, so $\angle AD'O=\angle D'AO$ implies $D'$ lies on the circumcircle of $ABC$.