lyukhson wrote: In a triangle $ABC$, let $D$ and $E$ be the feet of the angle bisectors of angles $A$ and $B$, respectively. A rhombus is inscribed into the quadrilateral $AEDB$ (all vertices of the rhombus lie on different sides of $AEDB$). Let $\varphi$ be the non-obtuse angle of the rhombus. Prove that $\varphi \le \max \{ \angle BAC, \angle ABC \}$. Let $d(A,p)$ be distance of point $A$ to line $p$. Lemma: For arbitrary point $M$ on straight line $DE$ holds: $d(M,AB)=d(M,AC)+d(M,BC)$ Proof: If $\frac{DM} {DE}=k$ holds $d(M,AB)=kd(E,AB)+(1-k)d(D,AB)=kd(E,BC)+(1-k)d(D,AC)=d(M,BC)+d(M,AC)$ which concludes the proof of lemma. Now let $KLMN$ be rhombus such $K$,$L$ and $N$ are on $AB$,$BD$ and $EA$ respectively. Let $a$ be side of rhombus and $O$ its circumcenter.We have: $d(M,AC)+d(M,BC)=a(\sin \angle MNC+\sin \angle MLC)$ and $d(M,AB)=2d(O,AB)=d(L,AB)+d(N,AB)=a(\sin \angle NKA+\sin \angle LKB)$. But if $\varphi>\angle BAC,\angle ABC$ then $\angle NKA=\angle MNC+\varphi - \angle BAC>\angle MNC$ and analoguosly $\angle LKB>\angle MLC$ so from above equations we get $d(M,AB)>d(M,AC)+d(M,BC)$ which is a contradiction. This concludes the proof.

Tricky. I think the main idea here is to disregard the inequality altogether and look at the structure. lyukhson wrote: In a triangle $ABC$, let $D$ and $E$ be the feet of the angle bisectors of angles $A$ and $B$, respectively. A rhombus is inscribed into the quadrilateral $AEDB$ (all vertices of the rhombus lie on different sides of $AEDB$). Let $\varphi$ be the non-obtuse angle of the rhombus. Prove that $\varphi \le \max \{ \angle BAC, \angle ABC \}$. Assume $\angle A, \angle B \le 90^{\circ}$ else we're done. Let $XZYT$ be the rhombus with $X, Y$ on $\overline{CA}, \overline{CB}$ and $Z, T$ on $\overline{AB}, \overline{DE}$ respectively. Suppose it has side length $1$. Let $O$ be the center of the rhombus. Claim. $d(T, AB)=d(T, CA)+d(T, CB)$ for any point $T$ on line $\overline{DE}$. (Proof) Clear for $T \in \{D,E\}$. Conclude by linearity. $\blacksquare$ Now let $\alpha=\angle TXC, \beta=\angle TYC, \gamma=\angle XZA, \delta=\angle YZB, \phi=\angle ZXT$. Then $$d(X, AB)+d(Y, AB)=2d(O, AB)=d(T, AB)=d(T, CA)+d(T, CB)$$hence $$\sin \alpha+\sin \beta=\sin \gamma+\sin \delta \iff \sin \alpha+\sin \beta=\sin(\alpha+\phi-A)+\sin(\beta+\phi-B)$$failing to hold if $\phi>A$ and $\phi>B$. $\blacksquare$ EDIT: okay, it is equivalent to the solution above

Here is a slightly less clever solution: A-index everything. WLOG let $\angle B\geq \angle C,$ and let the rhombus be $UVXY$ with center $O$ as shown. Assume $\triangle ABC$ is acute, as otherwise we are done. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 0, xmax = 6, ymin = -0.3602857286763528, ymax = 5.379315803770327; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pair A = (2.48,4.86), B = (1.18,0.3), C = (5.44,0.34), D = (4.039163451791813,2.4791152695611505), I = (2.966749270371574,1.6617733590935135), U = (2.9340461842372827,2.3975376779262008), X = (3.191815237929462,0.31889028392422025), O = (3.0629307110833723,1.3582139809252105), V = (1.4237329602498232,1.1549402297993807), Y = (4.64474751887375,1.5543720319900853); draw(A--B--C--cycle, linewidth(.4) + zzttqq); /* draw figures */ draw(A--B, linewidth(.4) + zzttqq); draw(B--C, linewidth(.4) + zzttqq); draw(C--A, linewidth(.4) + zzttqq); draw((xmin, 0.7621513447213091*xmin-0.5993385867711447)--(xmax, 0.7621513447213091*xmax-0.5993385867711447), linewidth(.4)); /* line */ draw((xmin, -0.5344275625836337*xmin + 3.2472859404549674)--(xmax, -0.5344275625836337*xmax + 3.2472859404549674), linewidth(.4)); /* line */ draw((xmin, 0.07381804088128033*xmin + 2.1809521367506086)--(xmax, 0.07381804088128033*xmax + 2.1809521367506086), linewidth(.4)); /* line */ draw((xmin, -8.063991251968687*xmin + 26.057660440487755)--(xmax, -8.063991251968687*xmax + 26.057660440487755), linewidth(.4)); /* line */ draw((xmin, 0.12400807103503081*xmin + 0.9783858517298062)--(xmax, 0.12400807103503081*xmax + 0.9783858517298062), linewidth(.4)); /* line */ /* dots and labels */ dot(A,dotstyle); label("$A$", (2.5060441331120797,4.9239954587382435), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (1.2042127240766882,0.3643790458113285), NE * labelscalefactor); dot(C,dotstyle); label("$C$", (5.468832857123662,0.4028568214478426), NE * labelscalefactor); dot(D,linewidth(4.pt) + dotstyle); label("$D$", (4.064394046390898,2.531960406668287), NE * labelscalefactor); dot((1.753130310568481,2.310364781686365),linewidth(4.pt) + dotstyle); label("$E$", (1.7813793586243987,2.3588104163039736), NE * labelscalefactor); dot(I,linewidth(4.pt) + dotstyle); label("$I$", (2.993429291174591,1.7111011930893203), NE * labelscalefactor); dot(U,dotstyle); label("$U$", (2.9613644781441626,2.4614178180013444), NE * labelscalefactor); dot(X,dotstyle); label("$X$", (3.21788298238759,0.3836179336295856), NE * labelscalefactor); dot(O,linewidth(4.pt) + dotstyle); label("$O$", (3.0896237302658762,1.4096919506032937), NE * labelscalefactor); dot(V,linewidth(4.pt) + dotstyle); label("$V$", (1.4479053031079439,1.204477147208552), NE * labelscalefactor); dot(Y,linewidth(4.pt) + dotstyle); label("$Y$", (4.667212531362952,1.6084937913919495), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Lemma: $BC>CD.$ Proof: By the Angle Bisector Theorem, $AD=kBA$ and $CD=kBC.$ Now $k<1$ by the Triangle Inequality. Lemma: $O$ lies in the exterior of $\triangle IDE.$ Proof: Assume otherwise, and let $S$ be the set of midpoints formed by $UX$ where $U$ varies on $DE$ and $X$ varies on $BC.$ We will show that all points in $S$ are lower than $I$ relative to $\overline{BC}$. Since $\angle B\geq \angle C,$ the highest point in $S$ is formed when $U=D$ and has a height of $\tfrac{\delta(D,BC)}{2}.$ But \[\frac{\delta(I,BC)}{\delta(D,BC)}=\frac{BI}{BD}=\frac{BI}{BI+ID}=\frac{BC}{BC+CD}>\frac{1}{2}\]by the previous lemma, so we are done. Now since $O$ doesn't lie in $\triangle IDE,$ it lies below at least one of $BID$ or $CIE,$ WLOG the first. Then if $V_1$ is the foot of $V$ onto $BC,$ $V_1$ lies on segment $BC$ (since $\triangle ABC$ is acute) and $VOXV_1$ is cyclic and so \[\frac{\varphi}{2}=\angle OVX=\angle OV_1X<\angle OBX<\angle IBX=\frac{\angle B}{2},\]completing the proof.

Generic_Username wrote: $\angle OV_1X<\angle OBX$ This part is wrong

Note that if $\max(\angle A,\angle B)>90$ we are already done, so suppose they are not obtuse. Let the rhombus have vertices $WXYZ$ on edges $AE,ED,DB,BA$ respectively. Note that the trilinear equation of line $DE$ is $x+y=z$. Furthermore, since $XTYZ$ is a parallelogram, we have that the sum of the distances from $X,Y$ to $AB$ is equal to the distance from $Z$ to $AB$. Hence, if we consider the feet $X',Y',Z_A,Z_B$ of the perpendiculars from $X,Y,Z,Z$ onto $AB,AB,AC,BC$, then we have $ZZ_A+ZZ_B=XX'+YY'$. Dividing by the side length of the rhombus, $\sin ATX+\sin YTB=\sin CXZ+\sin ZYC$, so WLOG $\sin ATX\le \sin CXZ$. The only way this can happen is if $\angle ATX\le \angle CXZ$, in which case $\angle ZXT\le \angle A$ as desired, or $\angle ATX\ge 180-\sin CXZ$, which can only occur when $(T,X)=(A,E)$, in which case $\angle XTY=\angle A$.

Any motivation for the first claim or it is just well-known? I tried this problem for a few hours before getting the first claim as a hint, then I finished the problem within half an hour. Let the rhombus be $MNOP$ with $M,N,O,P$ on $DE,EA,AB,BD$, respectively. In addition, let $\alpha$ and $\beta$ denote angles $MNC$ and $MPA$, respectively. Finally, let $s$ be the side length of the rhombus. The key claim is the following: Claim: The distance from $M$ to $AB$ is equal to the sum of the distances from $M$ to $AC$ and $BC$. Proof. Let $S$ be the set of all points following this condition. Notice that the condition in the claim is equivalent to $ax+by+cz=0$ in barycentric coordinates for constants $a,b,c$ and thus is a line. Since $D$ and $E$ satisfy the condition, $S$ is precisely the line $DE$, and $M$ is on this line. $\blacksquare$ Observe that diagonal $MO$ of the rhombus is at least the distance from $M$ to $AB$, which is equal to the sum of the distances from $M$ to $AC$ and $M$ to $BC$, which is equal to $s(\sin \alpha + \sin \beta)$ WLOG let $\angle A \ge \angle B$. Suppose for the sake of contradiction that $\angle MNO > \angle A$. Then $\angle C + \alpha + \beta = \angle NMP < \angle B + \angle C$, so $\alpha+\beta < \angle B$. Since the sine function is concave over $[0,\frac{\pi}{2}]$ $\sin \alpha + \sin \beta < 2\sin \frac{B}{2} \le 2\sin \frac{A}{2}$. Therefore we must have that $\frac{1}{2}MO \ge s\sin \frac{A}{2}$. Letting $Q$ be the center of the rhombus we get that $\sin \frac{1}{2}\angle MON = \frac{MQ}{MN} \ge \sin \frac{A}{2}$ and thus $\angle MON > \angle A$, contradiction.

anantmudgal09 wrote: $$\sin \alpha+\sin \beta=\sin \gamma+\sin \delta \iff \sin \alpha+\sin \beta=\sin(\alpha+\phi-A)+\sin(\beta+\phi-B)$$failing to hold if $\phi>A$ and $\phi>B$ This needs $\gamma$ and $\delta$ to be acute too, right? The case where one of them is obtuse is tricky to resolve and took me quite a while.

Case $\max \{ \angle BAC, \angle ABC \}\ \geq 90^\circ$ is trivial so suppose the opposite. Claim. For every point $X$ on segment $DE$ holds $d(X,AB)=d(X,AC)+d(X,BC).$ Proof. This clearly works for cases $X=D,E,$ and so by linearity for every $X\text{ } \Box$ Denote the rhombus by $PQRS$ (where $P\in AB,Q\in BC$) and let $O=PR\cap QS,|PQ|=a.$ $$d(R,AC)+d(R,AC)=a(\sin \angle CBR+\sin \angle CSR)$$$$d(R,AB)=2d(O,AB)=d(Q,AB)+d(S,AB)=a(\sin \angle APS+\sin \angle BPQ).$$But $\angle CSR$ either is $\angle BAC+\angle APS-\varphi$ or $\angle BAC+\angle APS+\varphi -\pi,$ similarly for $\angle CQR.$ Now easy check shows that $d(R,AB)=d(R,AC)+d(R,BC)$ can't occur for $\varphi > \max \{ \angle BAC, \angle ABC \}.$ We are done.

This is the hardest G3 I've ever done. WLOG $\angle B \leq \angle A < 90^{\circ}$. Increase $\angle B$ to $\angle A$ and draw the new lines. By parallel lines, a homothety through $A$ sending the point on $DB$ to the new $D'B'$ must send the point on $DE$ to a point under $D'E'$ since if these points are $X, Y$ respectively then $\frac{AX}{AX'}$ where $X'$ is the projection through $A$ of $X$ to $D'E'$ is $\geq \frac{AE}{AE'}$ by parallel lines with ratio lemma and $\frac{AY}{AY'} \leq \frac{AE}{AE'}$ by ratio lemma. Now simply use coordinate bash on arbitrary isosceles triangles to finish for arbitrary rhombuses whose top point lies below the new $D'E'$, where each unique angle is determined by the height of the top point of the rhombus above $AB$.

Let $P$ be on $AE$, $Q$ be on $DE$, $R$ be on $CD$, and $S$ be on $AB$ such that $PQRS$ is a rhombus. Let $A_1B_1C_1$ be the $Q$-pedal triangle in $\triangle ABC$ with $A_1$ opposite $A$ etc. Let $P_1$ and $R_1$ be feet of the altitude of $P$ and $R$ onto $AB$. Since $Q$ is on $DE$, we can say that $QA_1+QB_1=QC_1$. On the other hand, since $PQRS$ is a rhombus, $QC_1=PP_1+RR_1$. Thus, one of the following is true: $QA_1\ge RR_1$ or $QB_1\ge PP_1$. Without loss of generality, let it be the former. We have \[QP\sin (\angle EPQ) = QA_1\ge RR_1 = PS\sin(\angle PSA)\]so $\angle EPQ\ge \angle PSA$. Since $\angle QPS+\angle EPQ = \angle PAS+\angle PSA$, we have $\angle QPS\le \angle A$ as desired.

Random Lemma that I saw somewhere suddently summoned lol WLOG that $\triangle ABC$ is acute or else the problem is trivial and WLOG $AC \le BC$. Now let $WXYZ$ be our parallelogram with $W,X,Y,Z$ in $ED,DB,BA,AE$ respectively and suppose FTSOC $\varphi>\angle BAC$ Now from linearity it's easy to check that point $W$ satisfies that $d(W,AB)=d(W,AC)+d(W,BC)$, but notice that if you let $\ell$ to be the side of the parallelogram then this equality after canceling $\ell$ and using LoS becomes $\sin(\angle CZW)+\sin(\angle CXW)=\sin(\angle ZYA)+\sin(\angle XYB)$. But from angle chasing we also have that: \[\angle ZYA=\angle CZY-\angle BAC=\angle CZW+\varphi-\angle BAC>\angle CZW \]\[\angle XYB=\angle CXY-\angle CBA=\angle CXW+\varphi-\angle CBA>\angle CXW \]Now clearly note that $\angle ZYA+\angle XYB=\varphi \le 90$ so each of these is acute and thus the other two angles $\angle CZW, \angle CXW$ are as well, but now from increasingness of $\sin(x)$ in the range $\left(0, \frac{\pi}{2} \right)$ this is a contradiction. Therefore $\varphi \le \text{max} \{\angle BAC, \angle ABC \}$ thus we are done .