3
$\begingroup$

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:

enter image description here

${AB} \parallel EC_2$ 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.

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

2 Answers 2

1
$\begingroup$

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: $DC_2=KC_2-KD=13-{24\over5}={41\over5}$.

$\endgroup$
-2
$\begingroup$

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 $

enter image description here

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .