The Missing Point

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July 13th 2011, 07:23 PM
aldrincabrera

The Missing Point

,.hello everyone,.,can u help me with this??

Given points (-3,5) , (-2,-3) , (3,-4). find the fourth point in the first quadrant so that the quadrilateral made is both circumscribable and inscribable.

I managed to start with the solution. using my 3 points, i had drawn a triangle, then i drew a circle around it in which the 3 points are tangent to the circle. next, i got the intersection of the perpendicular bisectors of the 3 sides and concluded that it was the center of the circle. so with the center, i have many points for my inscribable quadrilateral..now, i am having a hard time finding which point will make it both circumscribable and inscribable,.,please help me,,.thnx
July 13th 2011, 09:52 PM
abhishekkgp

Re: The Missing Point

Quote:

Originally Posted by aldrincabrera

,.hello everyone,.,can u help me with this??

Given points (-3,5) , (-2,-3) , (3,-4). find the fourth point in the first quadrant so that the quadrilateral made is both circumscribable and inscribable.

I managed to start with the solution. using my 3 points, i had drawn a triangle, then i drew a circle around it in which the 3 points are tangent to the circle. next, i got the intersection of the perpendicular bisectors of the 3 sides and concluded that it was the center of the circle. so with the center, i have many points for my inscribable quadrilateral..now, i am having a hard time finding which point will make it both circumscribable and inscribable,.,please help me,,.thnx

i couldn't solve it completely but this may help..
1) the fourth point will lie on the circumcircle of the triangle ((-3,5),(-2,-3),(3,-4)) as you pointed out too.
2) All the four angle bisectors of the angles of the quadrilateral meet at a point. this is the necessary and sufficient condition for a circle to be inscribed in the quadrilateral.
July 14th 2011, 05:04 AM
earboth

1 Attachment(s)

Re: The Missing Point

Quote:

Originally Posted by aldrincabrera

,.hello everyone,.,can u help me with this??

Given points (-3,5) , (-2,-3) , (3,-4). find the fourth point in the first quadrant so that the quadrilateral made is both circumscribable and inscribable.

I managed to start with the solution. using my 3 points, i had drawn a triangle, then i drew a circle around it in which the 3 points are tangent to the circle. next, i got the intersection of the perpendicular bisectors of the 3 sides and concluded that it was the center of the circle. so with the center, i have many points for my inscribable quadrilateral..now, i am having a hard time finding which point will make it both circumscribable and inscribable,.,please help me,,.thnx

The bad news first: I can't provide you with the complete solution ... (Worried)

Here is what I've done so far:

1. You probably have determined the equation of the circumscribed circle as:

$\left(x-\frac32\right)^2+\left(y-\frac32\right)^2=\frac{65}2$

2. All straight lines passing through (-3, 5) with the slope s have the equation

$y=s x +3s+5$

All straight lines passing through (3, -4) with the slope m have the equation

$y=mx -3m -4$

3. The point P of intersection of these two lines is the missing 4th vertex of the quadrilateral which has to be located on the circle-line of the circumscribed circle:

$P\left(\frac{3(m+s+3)}{m-s} , \frac{3s(2s+3)}{m-s}+6s+5\right)$

4. When I tried to determine the value of s or m the equations grew to a very ugly size. So I guess that I've made the wrong approach.

5. So I tried to determine the coordinates of the center M(k, h) of the incircle. M must have the same (perpendicular) distance to all 4 sides of the quadrilateral. But with this approach I have 2 more variables and the system of equations became quite extensive.

I've attached a sketch (to scale) of the quadrilateral so it is possible for you to check your results of your calculations.
July 15th 2011, 11:33 AM
bjhopper

Re: The Missing Point

Hello,
A (-3,5) B (-2,-3) C (3,-4) Find point D forming a quad inscribable and circumscribable.
The quad has to be a regular trapezoid.Point D is (7,3) which can be calculated by coordinate geometry
Equation BC y = -1/5x -17/5 Point slope
Equation AD parallel to BC y= -1/5x +22/5 Point slope
Midpoint of BC is ( 0.5,-3.5) slope diagram
Equation of perpendicular bisector of BC y=5x-30/5 Point slope
Using this equation and equation AD solve for x and y the midpoint M of AD
x=2 y=4
Since AM must equal MD the coordinates of D become ( 7,3) congruent slope
diagrams

bjh
July 15th 2011, 04:54 PM
aldrincabrera

Re: The Missing Point

,.,sirs,.,thank you so much,.,.ammm,.,.does this mean that the point making the quad circumscribable and inscribable is unique???
July 15th 2011, 11:57 PM
Opalg

Re: The Missing Point

Quote:

Originally Posted by aldrincabrera

Given points (-3,5) , (-2,-3) , (3,-4). find the fourth point in the first quadrant so that the quadrilateral made is both circumscribable and inscribable.

I managed to start with the solution. using my 3 points, i had drawn a triangle, then i drew a circle around it in which the 3 points are tangent to the circle. next, i got the intersection of the perpendicular bisectors of the 3 sides and concluded that it was the center of the circle. so with the center, i have many points for my inscribable quadrilateral..now, i am having a hard time finding which point will make it both circumscribable and inscribable,.,please help me,,.thnx

I think that this problem does not have a neat solution.

The lengths of the two given sides of the quadrilateral are $\sqrt{65}$ and $\sqrt{26}.$ The length of the third side of the triangle formed by the three given vertices is $\sqrt{117}.$ (Notice that the squares of these distances are all multiples of 13.) That third side is a diagonal of the qudrilateral, and you can use the formula that I gave in this thread to see that the lengths of the other two sides of the quadrilateral must be $\tfrac{13}6\bigl(\sqrt{10}+1\bigr) \pm \tfrac{\sqrt{13}}2\bigl(\sqrt5-\sqrt2\bigr).$ Further laborious calculations show that those two sides must intersect at the point (x,y) given by

$x = \tfrac1{18}\bigl(66+12\sqrt{10} + 8\sqrt{26} - 2\sqrt{65}\bigr) \approx 7.1452763,$
$y = \tfrac1{18}\bigl(53+8\sqrt{10} - 12\sqrt{26} + 3\sqrt{65}\bigr) \approx 2.2942625.$

Those numerical values are consistent with the position of the point P in earboth's diagram, so I am fairly confident that my answer is correct. But it isn't nice. (Puke)

Edit. The point (7,3) proposed by bjhopper certainly lies on the circumcircle. But the regular trapezium (or trapezoid, to use the American terminology) formed in that way is not inscribable. In fact, the incentre would have to be at the midpoint of the line joining the midpoints of the two parallel sides. You can check that this is the point (5/4,1/4). The distance from this point to the parallel sides is $\tfrac34\sqrt{26}\approx3.82.$ But its distance from the sloping sides is $\tfrac9{20}\sqrt{65}\approx3.63.$ These are not the same, so that quadrilateral is not inscribable.

Second edit. The condition for a quadrilateral to be inscribable is that the sum of the lengths of each pair of opposite sides should be the same. The sides of bjhopper's quadrilateral have lengths $\sqrt{26}$ , $\sqrt{65}$ , $\sqrt{104}$ , $\sqrt{65}.$ It is certainly not true that $\sqrt{26} + \sqrt{104} = 2\sqrt{65}.$
July 16th 2011, 01:17 AM
aldrincabrera

Re: The Missing Point

.,.sir,.,thank u so much,.,ammm,.,.but any hint on how u find the point of intersection of the two sides??
July 16th 2011, 01:22 AM
aldrincabrera

Re: The Missing Point

,.,and sir,.,.i used the formula u gave me but im stuck with two unknowns,.,how should i cope with it??thnx sir
July 16th 2011, 05:32 AM
Opalg

Re: The Missing Point

Quote:

Originally Posted by aldrincabrera

any hint on how u find the point of intersection of the two sides??

You know that the point of intersection, P, must lie on the circumcircle, whose equation is $x^2+y^2-3x-3y = 28.$ Suppose that you know the length, d, of the side joining the point (–3,5) to P. Then P must lie on the circle centred at (–3,5) with radius d. The equation of that circle is $x^2+y^2+3x-10y = d^2-34.$ Subtract one of those circle equations from the other to get a linear relation between x and y. Now do the same thing for the distance of the remaining side from the point (3,–4), to get another linear relation between x and y. Finally, solve those two linear equations to get the values of x and y.

Quote:

Originally Posted by aldrincabrera

i used the formula u gave me but im stuck with two unknowns,.,how should i cope with it??

You know that two of the sides have lengths $\sqrt{26}$ and $\sqrt{65}.$ Suppose that the lengths of the other two sides are c and d. You know that the sum of the lengths of each pair of opposite sides must be the same. Therefore $\sqrt{26}+c = \sqrt{65}+d.$ That gives you one equation between c and d. The formula for the diagonal gives you another equation between c and d.

In practice, I found it easiest to write $p = \tfrac12(c+d)$ and $q = \tfrac12(c-d).$ Then $q = \tfrac12\bigl(\sqrt{65} - \sqrt{26}\bigr),$ $c = p+q$ and $d = p-q.$ Now use the formula for the diagonal to find that $p = \tfrac{13}6\bigl(\sqrt{10}+1\bigr).$

After all that hard work, I strongly suspect that the intended answer to the problem is the one that bjhopper gave, and that the person who set the question failed to realise that this is the wrong answer.
July 23rd 2011, 03:50 AM
aldrincabrera

Re: The Missing Point

,.,thnx sir,.,can i ask one last question sir??how did u get p=(1/2)(c-d),,.thnx sir
July 26th 2011, 10:34 PM
aldrincabrera

Re: The Missing Point

,.i somehow managed to solve the problem sir using ur strategy,.and i am very thankful to u,..,the only thing that is a mystery to me sir is the p=(1/2)(c-d),.how was it done sir??