A very nice problem! Construct the point $T$ in the outside such that $\measuredangle{TAD}=72^{\circ}$ and $\measuredangle{TDA}=\measuredangle{ABC}$. This means that $T, A, B$ are collinear and $\triangle{TAD}\sim\triangle{CAB}$. It results that $\frac{TA}{AC}=\frac{DA}{AB}\Longrightarrow \frac{TA}{DA}=\frac{CA}{BA}$. And because $\measuredangle{TAC}=\measuredangle{DAB}$, from the last relation we obtain $\triangle{TAC}\sim\triangle{DAB}.$ Continuing, let the lines $TC$ and $BD$ intersect in point $O$. Because triangles $TAC$ and $DAB$ are similar, we obtain $\measuredangle{ABO}=\measuredangle{ACO}$, thus the quadrilateral $AOCB$ is cyclic. This means that $\measuredangle{OAC}=\measuredangle{OBC}=18^{\circ}\Longrightarrow \measuredangle{DAO}=18^{\circ}$. Also, let $\measuredangle{ADB}=x\Longrightarrow \measuredangle{ABD}=72^{\circ}-x\Longrightarrow\measuredangle{OCA}=72^{\circ}-x$. Also, in cyclic quadrilateral $OABC$, we have $\measuredangle{OAB}=90^{\circ}\Longrightarrow \measuredangle{OCB}=90^{\circ}$. But $\measuredangle{DCB}=126^{\circ}\Longrightarrow \measuredangle{OCD}=36^{\circ}.$ Now we have calculated all the angles around point $O$ and the triangle $DAC$, thus from the generalised Ceva trig, we obtain \[\frac{\sin\measuredangle{DAO}}{\sin\measuredangle{OAC}}\cdot\frac{\sin\measuredangle{ACO}}{\sin\measuredangle{OCD}}\cdot\frac{\sin\measuredangle{ODC}}{\sin\measuredangle{ODA}}=1\Longrightarrow \frac{\sin18^{\circ}}{\sin18^{\circ}}\cdot\frac{\sin(72^{\circ}-x)}{\sin36^{\circ}}{\cdot\frac{\sin36^{\circ}}{\sin x}}=1 \] \[\Longrightarrow \sin (72^{\circ}-x)=\sin x\Longrightarrow x=72^{\circ}-x\Longrightarrow x=36^{\circ}\] This means that $\measuredangle{ADP}=36^{\circ}\Longrightarrow \measuredangle{APD}=108^{\circ}$!!

very nice solution andy but you've got a little typo at the very end : APD is 108. very nice solution ...

regular 10-agon

note that $\angle BCD=180-18-36=126$. now $\angle BCD+\frac{1}{2}\angle BAD=180$. also since $\angle DAC=2\angle DBC$ and $\angle BAC=2\angle BDC$, we conclude that the circle with center at $A$ of radius $AC$ passes through $B$ and $D$. so $AB=AC=AD$. hence $\angle ABD=\angle ADB=36$ and thus $\angle APD=72+36=108$.

this is a little similar to andyciup solution. let $\angle ADB=\alpha$. then, $\angle ABD=72-\alpha, \angle ACB=18+\alpha, \angle ACD=108-\alpha$. apply trig ceva to the quad ABCD, then, $\frac{\sin 36 \sin 36 \sin (72-\alpha) \sin (18+\alpha)}{\sin \alpha\sin 72\sin 18\sin (108-\alpha)}=1$. now, $\frac{\sin 36\sin 36}{\sin 72\sin 18}=2\sin 36$ and $\sin (72-\alpha) \sin (18+\alpha)=\frac{1}{2}\sin (36+2\alpha)$. this implies that $\frac{\sin 36\sin (36+2\alpha)}{\sin \alpha \sin (72+\alpha)}=1$. clearing the denominator and applying prod-to-sum formulas it follows that $\cos 2\alpha-\cos(72+2\alpha)=\cos 72-\cos (72+2\alpha)$ or $\cos 2\alpha=\cos 72$. since $\alpha<72<90$ it follows that $2\alpha=72$ or $\angle APD=180-36-\alpha=108$. btw, nice problem, and it seems a little similar in its form to bmo 07 p1. is the author of this problem the same as the one from the bmo?

Let $E$ be the point on $AC$, different from $P$, such that $\angle EBC = 18$. Then $\angle DAE = \angle DBE = 36$, then ABED is cyclic and $\angle EDB = \angle EAB = 72$. Now $\angle EDC = \angle BDC = 36$, therefore C is the incenter of BED, and $\angle BEC = \angle DEC = 36$. So, $\angle ABD = \angle AED = 36$ and $\angle APD = \angle PAB+\angle PBA = 72+36 = 108$.

pohoatza wrote: Let $ ABCD$ be a convex quadrilateral with $ \angle{DAC}= \angle{BDC}= 36^\circ$ , $ \angle{CBD}= 18^\circ$ and $ \angle{BAC}= 72^\circ$. Determine the measure of $ \angle{APD}$, where $ P\in AC\cap BD$. Quote: Lemma (extension). Let $ ABCD$ be a convex quadrilateral. Prove that $ \{\begin{array}{c}m(\widehat{CBD})=x\ ,\ m(\widehat{BDC})=2x\\\ m(\widehat{CAD})=y\ ,\ m(\widehat{CAB})=2y\\\ P\in AC\cap BD\ ,\ m(\widehat{APD})=z\end{array}$ $ \implies$ $ \boxed{\cos (z+x-y)+\cos (z+y-x)+\cos (x+y-z)=0}\ \ (*)$.

Remark. The relation $ (*)$ is symmetrically in the variables $ x$, $ y$. Thus, I can propose the following problem : Quote: Let $ ABCD$ be a convex quadrilateral with $ \angle{DAC}=18^{\circ}$, $ \angle{BDC}= 72^\circ$ and $ \angle{CBD}=\angle{BAC}= 36^\circ$. Determine the measure of $ \angle{APD}$, where $ P\in AC\cap BD$. Proof of the proposed problem. For $ \{\begin{array}{c}x=18^{\circ}\\\ y=36^{\circ}\end{array}$ the relation from the above lemma becomes $ \cos (54-z)+\cos (z-18)+\cos (z+18)=0$, i.e. $ f(z)\equiv\sin (36+z)+2\cos z\cdot\cos 18=0$. Observe that $ z>90^{\circ}$ and on the interval $ (\frac{\pi}{2},\pi )$ the function $ f$ is strict decreasing and $ f(108^{\circ})=0$. In conclusion, $ m(\widehat{APD})=108^{\circ}$.

We prove the circumcenter O of triangle BCD superimposes with A, ffrom where will find angle APD=108. Suppose the contrary. DOC=DAC=36 so AOCD is cyclic. COB=CAB=72 so AOBC is cyclic as well. From cyclic AOCD we find CDO=CAO=72=CAB so O lies on AB and lies on the circumscribed circle of triangle ABC so O=A or O=B. But O is circumcenter of BCD so O cannot superimpose with B.

Proof extend $ CA$ to point $ E$ such that $ AE=AB$ then $ \angle{AEB}=\angle{ABE}= 36^{\circ}=\angle{CDB}$ then $ CBED$ is concyclic. so $ \angle{CED}=\angle{CBD}= 18^{\circ}$ and $ \angle{ADE}=\angle{DAC}-\angle{CED}= 36^{\circ}-18^{\circ}=18^{\circ}$ which mean $ AD=AE=AB$. hence $ A$ is circumcenter of quadrilateral $ BCDE$ then $ AD=AC$ and $ \angle{ACD}=72^{\circ}$ $ \angle{APD}=\angle{ACD}+\angle{BDC}=36^{\circ}+72^{\circ}=\boxed{108^{\circ}}$ Tummy

Same solution of Nayel. One question though, just because the angles are twice as much can we conclude that A is the center of the circle?

Yeah, simple solution. We know that, the circumcircle of $\triangle BCD$ have center at point $A$. So, $\triangle ABD$ is isosceles triangle. So, $\angle ABD = \frac{180^{\circ} - 108^{\circ}}{2} = 36^{\circ}$. So, $APD = 180^{\circ} - 36^{\circ} - 36^{\circ} = \boxed{108^{\circ}}$

@ youarebad, I am not sure your claim is obvious (you need either $AC=AD$, or $AB=AC$ as well), but: If the interior angle bisector of $\angle BAC$ intersects $BD$ at $E$, then from $\angle EAC=\angle EDC=36^\circ, AECD$ is cyclic and $CE=CD$ (from $\angle CAE=\angle DAC$), so $\angle CED=\angle CAD=36^\circ$, consequently $\angle BCE=\angle CBE=18^\circ$ and $BE=CE$; $E$ being inside the triangle $\triangle ABC$, lying on the interior angle bisector of $\angle BAC$ and on the perpendicular bisector of $BC$, the triangle $ABC$ is isosceles, so $\angle BEC=180^\circ-36^\circ=144^\circ=2\cdot m(\angle BAC)$ implies $E$ is circumcenter of $\triangle ABC$, hence $AE=CE$; from $AECD$ cyclic, we get $BD$ angle bisector of $\angle ADC\implies \angle ADB=36^\circ$; with $\angle CAD=36^\circ$, we get the desired result, $\angle APD=108^\circ$. Best regards, sunken rock

youarebad wrote: Yeah, simple solution. We know that, the circumcircle of $\triangle BCD$ have center at point $A$. So, $\triangle ABD$ is isosceles triangle. So, $\angle ABD = \frac{180^{\circ} - 108^{\circ}}{2} = 36^{\circ}$. So, $APD = 180^{\circ} - 36^{\circ} - 36^{\circ} = \boxed{108^{\circ}}$ Imho the bulk of the problem is proving your assumption.

The rest is trivial.

Let $C_{1}$ be the circumcircle of triangle BCD . Let E be the point where AC intersects with $C_{1}$ . The quadrilateral BCDE is cyclic , so we have $<DEC=<DBC=18^{\circ}$ and $<CEB=<BDC=36^{\circ}$ Aso we have $<DAE=<ADE+<DEA $<=> $<ADE=18^{\circ}$ therefore triangle ADE is isosceles . Similar we get triangle ABE isosceles and we conclude that A is te circumcenter of triangle BDE and also of triangle BDC (because BCDE is cyclic) . Hence we have that triangle ADC is isosceles and $<ADC=<ACD$ <=> $<ADB=36^{\circ}$ and we have that $<APD=108^{\circ}$

nayel wrote: note that $\angle BCD=180-18-36=126$. now $\angle BCD+\frac{1}{2}\angle BAD=180$. also since $\angle DAC=2\angle DBC$ and $\angle BAC=2\angle BDC$, we conclude that the circle with center at $A$ of radius $AC$ passes through $B$ and $D$. so $AB=AC=AD$. hence $\angle ABD=\angle ADB=36$ and thus $\angle APD=72+36=108$. NICE ONE!!!!

Solution: Construction: Let $X$ be on $BD$ such that $ADCX$ is cyclic. Let $Y$ be on $BD$ such that $BCYA$ is cyclic. Claim: $\angle BCY = 90^{\circ}$. Proof: $\angle XCB = \angle DXC - 18^{\circ} = \angle DAC - 18^{\circ} = 18^{\circ}$. $\angle XCY = 180^{\circ} - 36^{\circ} - \angle XYC = 144^{\circ} - 72^{\circ} = 72^{\circ}$. So, $\angle BCY=90^{\circ}$. Claim: $YX=XB=AX$ Proof: $YX=XB$ is trivial by the above angle chase. As $\angle YAB = 90^{\circ}$, $AX$ is the median. So, $YX=XB=AX$. So, $\angle APD = 180^{\circ} - 18^{\circ} - (18+36)^{\circ} = 108^{\circ}$. $\blacksquare$

Let the circle $\odot (ABD)$ intersect $AC$ second time at $E$. It follows that $\angle BDE=72^\circ$ and $\angle DBE=36^\circ$, i.e. $C$ is the incenter of $\triangle BED$ and $AC$ the bisector of $\angle BED$, hence $AD=AB$, making the required angle $\angle APD=108^\circ$. Best regards, sunken rock

Someone please check this as my solution does not incorporate the fact that $\angle CBD=18^\circ.$

I think this is a different solution

btw @(someone_i_know_but_u_dont) that IMO 2017 P4 was too good oof, i reflected almost everything across $S$ except $A$ i was so high and also ggb is too bad for angle chase, too hard to write angles on screen, ig have to switch to pen and paper sadge

proshi wrote: Someone please check this as my solution does not incorporate the fact that $\angle CBD=18^\circ.$

If I am not mistaken, $\triangle DPA \sim \triangle CPQ$ does not imply $DA \parallel CQ$.

Let $O$ be the circumcenter of triangle $BCD$; clearly $O\not\equiv D$. We compute $\angle BCD = 180^{\circ} - \angle DBC - \angle BDC = 126^{\circ}$ and $\angle BOD = 360^{\circ} - 2\angle BCD = 108^{\circ}$, as well as $\angle DOC = 2\angle DBC = 36^{\circ}$. Hence $\angle DOC = \angle DAC$ and $\angle BAD = \angle BOD$ and if we suppose that $O\not\equiv A$, then it would follow that the circumcircles of triangles $ACD$ and $ABD$ have $O$ as a third common intersection point, contradiction. Therefore $O\equiv A$, hence $AC = AD$, thus $\angle ACD = 90^{\circ} - \frac{\angle DAC}{2} = 72^{\circ}$ and $\angle APD = \angle PCD + \angle PDC = 108^{\circ}$.

$A$ is circumcenter of $\triangle BCD$, the rest just angle chase!