kyotsu-test 2018 QCourse1-IV

kyotsu-test · Japan · eju-math__session1 Sine and Cosine Rules Multi-step composite figure problem
The triangle ABC satisfies
$$\mathrm { AB } = 4 , \quad \mathrm { AC } = 3 \quad \text { and } \quad \angle \mathrm { B } = 30 ^ { \circ } .$$
D is the point on side BC such that $\mathrm { AC } = \mathrm { AD }$. Let us consider the circumscribed circle O of triangle ACD.
(1) Since $\sin B = \frac { \mathbf { A } } { \mathbf { B } }$, we have $\sin C = \frac { \mathbf { C } } { \mathbf{D} }$.
Hence the radius of circle O is $\frac { \mathbf { E } } { \mathbf{F} }$.
(2) We have
$$\mathrm { BC } = \mathrm { G } \sqrt { \mathrm { H } } + \sqrt { \mathrm{H} }$$
and
$$\mathrm { BD } = \mathrm { J } \sqrt { \mathrm {~K} } - \sqrt { \mathrm { K } } .$$
Let us denote the intersection of side AB and circle O by E . Then
$$\mathrm { BE } = \frac { \mathbf { M } } { \mathbf{N} } .$$
Hence the relationships between the areas of triangles $\mathrm { BDE } , \mathrm { ADE }$ and ACD are
$$\begin{aligned} & \triangle \mathrm { BDE } : \triangle \mathrm { ADE } = \mathbf { O } : \mathbf { P } , \\ & \triangle \mathrm { BDE } : \triangle \mathrm { ACD } = \mathbf { Q } ( \mathbf{J} \sqrt { \mathbf { K } } - \sqrt { \mathrm { K } } ) : \mathbf { R S } \sqrt { \mathbf { T } } . \end{aligned}$$
The triangle ABC satisfies

$$\mathrm { AB } = 4 , \quad \mathrm { AC } = 3 \quad \text { and } \quad \angle \mathrm { B } = 30 ^ { \circ } .$$

D is the point on side BC such that $\mathrm { AC } = \mathrm { AD }$. Let us consider the circumscribed circle O of triangle ACD.

(1) Since $\sin B = \frac { \mathbf { A } } { \mathbf { B } }$, we have $\sin C = \frac { \mathbf { C } } { \mathbf{D} }$.

Hence the radius of circle O is $\frac { \mathbf { E } } { \mathbf{F} }$.\\
(2) We have

$$\mathrm { BC } = \mathrm { G } \sqrt { \mathrm { H } } + \sqrt { \mathrm{H} }$$

and

$$\mathrm { BD } = \mathrm { J } \sqrt { \mathrm {~K} } - \sqrt { \mathrm { K } } .$$

Let us denote the intersection of side AB and circle O by E . Then

$$\mathrm { BE } = \frac { \mathbf { M } } { \mathbf{N} } .$$

Hence the relationships between the areas of triangles $\mathrm { BDE } , \mathrm { ADE }$ and ACD are

$$\begin{aligned}
& \triangle \mathrm { BDE } : \triangle \mathrm { ADE } = \mathbf { O } : \mathbf { P } , \\
& \triangle \mathrm { BDE } : \triangle \mathrm { ACD } = \mathbf { Q } ( \mathbf{J} \sqrt { \mathbf { K } } - \sqrt { \mathrm { K } } ) : \mathbf { R S } \sqrt { \mathbf { T } } .
\end{aligned}$$