A geometry problem by Anish Roy

Geometry Level 3

In triangle $ABC$ , $AB=6, AC=10, BC=8$ . $BD$ is perpendicular to $AC$ . A circle is constructed with centre $B$ and radius is $BD$ such that circle intersect $AB$ at $P$ and $BC$ at $Q$ . Find $AP:QC$ .

3:5

1:2

2:5

3:8

1 solution

Dan Ley
Dec 3, 2016

From the question and the diagram:

$x+y=10$

$x^2+r^2=6^2$

$y^2+r^2=8^2$

$\implies y^2-x^2=64-36=28$

$\implies (y-x)(y+x)=28$

$\implies 10(y-x)=28$

$\implies y-x=2.8$

$\implies x=3.6, \space r=4.8$

Thus, $AP=6-4.8=1.2$ and $CQ=8-4.8=3.2$

$AP:CQ=1.2:3.2=3:8$

@Dan Ley using too much equations. You can find the value of r (in your picture) by using the formula Area of a triangle = 1/2 * b * h = 1/2 * hyp. * radius. = 1/2 * 10 *radius

Vishwash Kumar ΓΞΩ - 4 years, 6 months ago

Ahh, I didn't spot the Pythagorean triple! At least my method works for non-right triangles:)

Dan Ley - 4 years, 6 months ago

but you must hunt for the easiest way

Vishwash Kumar ΓΞΩ - 4 years, 6 months ago

@Vishwash Kumar Γξω – True! Areas can often provide useful shortcuts, like energy does in mechanics.

Dan Ley - 4 years, 6 months ago

@Dan Ley – yes but i am illiterate when it comes to classical mechanics LOLOL

Vishwash Kumar ΓΞΩ - 4 years, 6 months ago

@Dan Ley – so don't know much about that..........................................................................

Vishwash Kumar ΓΞΩ - 4 years, 6 months ago

although that too is right Nice LaTeX

Vishwash Kumar ΓΞΩ - 4 years, 6 months ago

@Dan Ley Are you new here

Vishwash Kumar ΓΞΩ - 4 years, 6 months ago

Relatively, been on a year, you?:)

Dan Ley - 4 years, 6 months ago

I am just two month old here

Vishwash Kumar ΓΞΩ - 4 years, 6 months ago

@Vishwash Kumar Γξω – That's cute;)

Dan Ley - 4 years, 6 months ago

A geometry problem by Anish Roy

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