# Find the largest possible area of $\triangle ABC. A(6,0), B(0,8)), C(x,y)$ where $x^2-16x+y^2-12y+91=0$

Taken from Singapore Mathematical Olympiad (SMO) senior 2019 Question 9

The coordinates of the vertices of a triangle $$\triangle ABC$$ are $$A(6,0), B(0,8)$$ and $$C(x,y)$$ such that $$x^2-16x+y^2-12y+91=0$$. Find the largest possible value of the area of the triangle $$\triangle ABC$$

my attempt: Since $$C$$ lies on a circle of centre $$(8,6)$$ and radius $$3$$, I constructed the following diagram:

$${AB} \parallel EC_2$$ and $$EC_2$$ is tangent to circle $$O$$. By tangent $$\perp$$ radius, $$\angle EC_2O = 90^{\circ}$$. Thus by corresponding angles, $$\angle C_2DA = 90^{\circ}$$

acute angle of line $$AB$$ with x-axis $$=-tan^{-1}(-8/6)=tan^{-1}(4/3)$$

acute angle of line $$OA$$ with x-axis $$=tan^{-1}(6/(8-6))=tan^{-1}(3)$$

$$\angle BAO = 180^{\circ} - tan^{-1}(4/3) - tan^{-1}(3) =...= tan^{-1}(13/9)$$ (by arctangent addition formula)

$$|OA| = \sqrt{6^2+(8-6)^2}=\sqrt40$$

$$|OD| = \sqrt {40} sin(tan^{-1}(13/9)) =...= 26/5$$

$$|AB| = \sqrt{(6)^2+(8^2)} = 10$$

$$|OC_2| = 3$$ (radius of circle $$O$$)

Since maximum area of $$\triangle ABC$$ would be when $$C$$ is in position $$C_2$$, area = $$0.5(3+26/5)(10)=41$$

Is there any simplier/alternate way of solving this question? I think that there can be a shorter method to get the answer but I am unsure how to proceed.

• $S_{max}=\frac{1}{2}*10*(10+3-4.8)=41$. Aug 14, 2022 at 2:19
• If calculus is allowed, there are only two points on the circle where $\frac{dy}{dx}=-\frac{4}{3}$ Aug 14, 2022 at 2:23
• To avoid trig, just note that line $OD$ passes through the origin. Aug 14, 2022 at 7:18
• @Intelligentipauca That is true. With that method, I can find the point where the perpendicular line intersects $AB$, making the calculation easier. Aug 14, 2022 at 10:43

Set $$K=(0,0)$$ and $$H=(8,0)$$. Note that triangles $$ABK$$ and $$KOH$$ are congruent, hence $$OK\perp AB$$ and $$OKDC_2$$ are collinear.
It follows that $$KC_2=13$$ and the altitude from $$D$$ in triangle $$KDA$$ is $$h={12\over25}\cdot6$$, so that $$KD={5\over3}h={24\over5}$$.
Finally: $$DC_2=KC_2-KD=13-{24\over5}={41\over5}$$.
Since area is constant on a give base and given altitude you are right in constructing a parallel to $$AB$$ to the circle at maximum extreme tangent point. You are right on track finding the required red triangle.
The base and height are now known, $$A=\frac12\cdot AB\cdot h$$