Vrijeme: 07:38

Boris mi nije dao ime | Boris didn't name this one #1

Neka je ABC trokut u kome je AB=10, AC=11. Njemu upisana kružnica dodiruje BC,CA,AB u P,Q,R redom. Neka je T nožište visine iz P na QR. Ako je \angle BTC=90^{\circ} nađi BC. Rješenje zaokruži na 4 decimale.
Let ABC be a triangle in which AB=10, AC=11. The circle inscribed to it touches BC,CA,AB in P,Q,R respectively. Let T be the base of the height from P to QR. If \angle BTC=90^{\circ} find BC. Round the solution to 4 decimal places.