Let-ABCD-be-a-convex-quadrilateral-with-DAB-BDC-90-Let-the-incircles-of-triangles-ABD-and-BCD-touch-BD-at-P-and-Q-respectively-with-P-lying-in-between-B-and-Q-If-AD-999-and-PQ-200-then-

FacebookTweetPin

Question Number 19792 by Tinkutara last updated on 15/Aug/17

$$\mathrm{Let}ABCD\mathrm{be}\mathrm{a}\mathrm{convex}\mathrm{quadrilateral}$$$$\mathrm{with}\mathrm{\angle }DAB=\mathrm{\angle }BDC=90°.\mathrm{Let}\mathrm{the}$$$$\mathrm{incircles}\mathrm{of}\mathrm{triangles}ABD\mathrm{and}BCD$$$$\mathrm{touch}BD\mathrm{at}P\mathrm{and}Q,\mathrm{respectively},$$$$\mathrm{with}P\mathrm{lying}\mathrm{in}\mathrm{between}B\mathrm{and}Q.\mathrm{If}$$$$AD=999\mathrm{and}PQ=200\mathrm{then}\mathrm{what}\mathrm{is}$$$$\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{radii}\mathrm{of}\mathrm{the}\mathrm{incircles}\mathrm{of}$$$$\mathrm{triangles}ABD\mathrm{and}BDC?$$

Answered by ajfour last updated on 16/Aug/17

Commented by ajfour last updated on 16/Aug/17

$$\mathrm{DP}=\mathrm{rcot}\theta ;\mathrm{DQ}=\mathrm{R};\mathrm{QP}=\mathrm{d}$$$$\mathrm{AD}=\mathrm{a}=\mathrm{rcot}\theta +\mathrm{r}\dots \left(\mathrm{i}\right)$$$$\mathrm{As}\mathrm{DP}=\mathrm{DQ}+\mathrm{QP}$$$$\mathrm{rcot}\theta =\mathrm{R}+\mathrm{d}\dots \left(\mathrm{ii}\right)$$$$\mathrm{from}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)$$$$\mathrm{a}-\mathrm{r}=\mathrm{R}+\mathrm{d}$$$$\Rightarrow \mathbf{R}+\mathbf{r}=\mathbf{a}-\mathbf{d}.$$

Commented by Tinkutara last updated on 16/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

Answered by ajfour last updated on 16/Aug/17

$$\mathrm{R}+\mathrm{r}=\mathrm{AD}-\mathrm{PQ}.$$$$\mathrm{R}+\mathrm{r}=\mathrm{AD}-\mathrm{PQ}.$$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

FacebookTweetPin