Junior Balkan MO

Geometry Difficulty 5.2 AIME, harder Find the answer

Let $ABCD$ be a convex quadrilateral with $\angle{DAC}= \angle{BDC}= 36^\circ$ , $\angle{CBD}= 18^\circ$ and $\angle{BAC}= 72^\circ$ . The diagonals and intersect at point $P$ . Determine the measure of $\angle{APD}$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let I be the intersection between $(DP)$ and the angle bisector of $\angle{DAP}$ So $\angle{CAI}=\angle{PAI}=36/2°=18°$ So $\angle{CAI}=18°=\angle{CBD}=\angle{CBI}$ We can conclude that $A,B,C,I$ are on a same circle.
So $\angle{ICB}=180-\angle{IAB}=180-\angle{IAC}-\angle{CAB}=180-18-72=90$ Because $\angle{CBD}=18$ and $\angle{CDB}=36$ we have $126=\angle{DCB}=\angle{ICB}+\angle{ICD}=90+\angle{ICD}$ So $\angle{ICD}=36=\angle{BDC}=\angle{IDC}$ So $I$ is on the angle bisector of $\angle{DAP}$ and on the mediator of $DC$ .
The first posibility is that $I$ is the south pole of $A$ so $I$ is on the circle of $DAC$ but we can easily seen that's not possible
The second possibility is that $DAC$ is isosceles in $A$ . So because $\angle{DAC}=36$ and $DAC$ is isosceles in $A$ we have $\angle{ADC}=72$ . So $\angle{APD}=180-\angle{PAD}-\angle{PDA}=180-36-(\angle{ADC}-\angle{PDC})=180-36-72+36=108$

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.