<Solid-2026-270>, hint

 

<Solid-2026-270>, pyramid, angle

<Singapore MO 2003>

 

<Cir-2026-798>, hint

 

<Cir-2026-798>, circle, right triangle

<Singapore MO 2002>

 

<Qua-2026-812>, hint

Let <ABC=<ADC=X AB=a, AD=b 

Then <EAF=180-X-2(90-X)=X. AE/AF=a sinX/ b sinX=a/b

Hence Triangles AEF and ABC are similar, EF/AC=AE/AB=sinX=35/37

Let R be the circumradius of  AEF. Since AH=2RcosX, EF/sin<EAF=2R , 35/sinX=2R

2R=35/(35/37)=37. AH=37 cosX( sinX=35/37). Finally AH=12.