Geometry problems with a parameter

In a problem with a parameter the given facts are letters, such as $\displaystyle a$ and $\displaystyle \alpha$, and the answer is an expression in them. The way is the same as with numbers:

• Mark the given facts on the drawing, and express the other angles with them.
• Choose a triangle where enough is known, and use the law of sines, the law of cosines or the area.
• Simplify with identities. It is worth checking the answer at one value of the parameter.

In triangle $\displaystyle ABC$: $\displaystyle {AB}={c}$, $\displaystyle {AC}={b}$, $\displaystyle {\angle BAC}={\alpha}$. $\displaystyle AD$ bisects the angle $\displaystyle A$, and $\displaystyle D$ is on $\displaystyle BC$.
Express $\displaystyle AD$ with $\displaystyle b$, $\displaystyle c$ and $\displaystyle \alpha$.

The solution below writes the area of the triangle in two ways: whole, and as the sum of the two triangles that $\displaystyle AD$ cuts it into. Let $\displaystyle {AD}={l}$.

• $\displaystyle {S_{ABC}}={\frac{1}{2}bc\sin\alpha}$.
• $\displaystyle {S_{ABD}+S_{ADC}}={\frac{1}{2}cl\sin\frac{\alpha}{2}+\frac{1}{2}bl\sin\frac{\alpha}{2}}={\frac{1}{2}(b+c)\,l\sin\frac{\alpha}{2}}$.
• Compare: $\displaystyle {l}={\frac{bc\sin\alpha}{(b+c)\sin\frac{\alpha}{2}}}$.
• $\displaystyle {\sin\alpha}={2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}$, so $\displaystyle {l}={\frac{2bc\cos\frac{\alpha}{2}}{b+c}}$.

A check: $\displaystyle {b}={6}$, $\displaystyle {c}={9}$, $\displaystyle {\alpha}={60^\circ}$ give $\displaystyle {l}={\frac{2\cdot54\cdot\cos30^\circ}{15}\approx6.24}$.

An isosceles triangle $\displaystyle ABC$: $\displaystyle {AB}={AC}={b}$, and the apex angle is $\displaystyle {\angle BAC}={\alpha}$.

Express with $\displaystyle b$ and $\displaystyle \alpha$ the base $\displaystyle BC$, the area of the triangle and the radius of the circumscribed circle. A hint: the altitude to the base bisects the apex angle.

$\displaystyle ADC$ is an isosceles triangle: $\displaystyle {AD}={AC}$.
The point $\displaystyle B$ is on the side $\displaystyle DC$, with $\displaystyle {AB}={BC}$ and $\displaystyle {DC}={3BC}$.

Let $\displaystyle {BC}={x}$ and $\displaystyle {\gamma}={\angle ACD}$. Explain why $\displaystyle {AC}={2x\cos\gamma}$ (in triangle $\displaystyle ABC$) and why $\displaystyle {DC}={2AC\cos\gamma}$ (in triangle $\displaystyle ADC$). From these, $\displaystyle {3x}={4x\cos^2\gamma}$.
Let $\displaystyle {c}={\cos\gamma}$ and solve.

Which solution fits a base angle of a triangle? Find the angles of triangle $\displaystyle ADC$.

The area of triangle $\displaystyle ADC$ is also given: $\displaystyle 16\sqrt{3}$ cm². $\displaystyle BT$ is the altitude to the side $\displaystyle AC$ in triangle $\displaystyle ABC$.
Find $\displaystyle AC$, and then the length of $\displaystyle DT$. A hint: $\displaystyle T$ is the midpoint of $\displaystyle AC$ – why?