Let-ABCD-be-a-convex-quadrilateral-and-let-k-gt-0-be-a-real-number-Find-the-locus-of-points-M-in-its-interior-such-that-MAB-2-MCD-k-

Question Number 16066 by Tinkutara last updated on 21/Jun/17

Let A B C D be a convex quadrilateral

and let k > 0 be a real number . Find

the locus of points M in its interior

such that

[M A B] + 2 [M C D] = k .

Answered by mrW1 last updated on 28/Jun/17

Let AB = a, CD = b

\Rightarrow [MAB] = \frac{1}{2} {ah}_{1}

\Rightarrow [MCD] = \frac{1}{2} {bh}_{2}

with

h_{1} = distance from M to side AB

h_{2} = distance from M to side CD

[M A B] + 2 [M C D] = k = constant

\Rightarrow \frac{1}{2} ({ah}_{1} + {2 bh}_{2}) = k

\Rightarrow \frac{1}{2} ({ah}_{1} + b^{'} h_{2}) = k with b^{'} = 2 b

Commented by mrW1 last updated on 28/Jun/17

Commented by mrW1 last updated on 29/Jun/17

Case 1 : AB / / CD

h_{1} + h_{2} = h = constant

\frac{1}{2} ({ah}_{1} + b^{'} h_{2}) = k

\Rightarrow \frac{1}{2} [{ah}_{1} + b^{'} (h - h_{1})] = k

\Rightarrow (a - b^{'}) h_{1} + b^{'} h = 2 k

if a = b^{'} :

\Rightarrow b^{'} h = 2 k \Rightarrow k = \frac{b^{'} h}{2} = \frac{ah}{2}

i . e . k must be \frac{ah}{2}, otherwise there is

no solution . In this situation M can

be everywhere in interior of ABCD .

if a \neq b^{'} :

(a - b^{'}) h_{1} + b^{'} h = 2 k

\Rightarrow h_{1} = \frac{2 k - b^{'} h}{a - b^{'}} = (\frac{k}{\frac{b^{'} h}{2}} - 1) \times (\frac{1}{\frac{a}{b^{'}} - 1}) \times h = constant

\Rightarrow the locus of M is a straight line

parallel to AB and CD . But there

is a solution only if the value of k is

between \frac{ah}{2} and \frac{b^{'} h}{2}, see diagram above .

When k = \frac{ah}{2}, M lies on side CD .

When k = \frac{b^{'} h}{2}, M lies on side AB .

Commented by mrW1 last updated on 29/Jun/17

Case 2 : AB and CD meets at O

∠ AOD = α

\frac{1}{2} ({ah}_{1} + b^{'} h_{2}) = k

\Rightarrow {ah}_{1} + b^{'} h_{2} = 2 k

that means point M has to fulfill that

{ah}_{1} + b^{'} h_{2} = constant which is 2 k .

We make {OA}^{'} = AB = a and

{OD}^{'} = 2 CD = 2 b = b^{'} .

We get A^{'} D^{'} .

Every point on A^{'} D^{'} fulfills that

{ah}_{1} + b^{'} h_{2} = constant .

Every point on any straight line / / A^{'} D^{'}

fulfills also this condition .

Each line parallel to A^{'} D^{'} stands for

a certain value of k from 0 to + \infty .

We have following special lines / / A^{'} D^{'} :

A 1 corresponding to k_{1} = {2 A}_{Δ ACD}

D 2 corresponding to k_{2} = A_{Δ ADB}

C 3 corresponding to k_{3} = A_{Δ ACB}

B 4 corresponding to k_{4} = {2 A}_{Δ CBD}

So we have :

if k < k_{1} \Rightarrow no solution for point M

if k_{1} ⩽ k ⩽ k_{2} \Rightarrow Locus of M is a part of straight line from AD to AB

if k_{2} ⩽ k ⩽ k_{3} \Rightarrow Locus of M is a straight line EF / / A^{'} D^{'}

if k_{3} ⩽ k ⩽ k_{4} \Rightarrow Locus of M is a part of straight line from BC to AB

if k > k_{4} \Rightarrow no solution for point M

Commented by Tinkutara last updated on 29/Jun/17

Thanks Sir! Very much .

Commented by mrW1 last updated on 28/Jun/17

Commented by mrW1 last updated on 28/Jun/17

Here is how to determine the line EF

for a certain value of k .

Area of Δ {OA}^{'} D^{'} = A_{0}

A_{0} = {ab}^{'} \sin α

A^{'} D^{'} stands for k = k_{0} = A_{0}

EF stands for k

\frac{OE}{{OA}^{'}} = \sqrt{\frac{k}{A_{0}}}

\Rightarrow OE = \sqrt{\frac{k}{A_{0}}} a

Answered by Tinkutara last updated on 29/Jun/17

Combining Problems 3 and 5 and using

the construction above, we deduce

that the area of X Y^{'} M is constant .

(Figure) . It results that the locus is

(according to the value of k) either a

segment K L parallel to X Y^{'}, or a

point, or the empty set .

Commented by Tinkutara last updated on 29/Jun/17

Commented by Tinkutara last updated on 29/Jun/17

This is my {book}^{'} s solution . Here

Problem 3 is Q . 16064 and 5 is Q . 16066

problem . (This Question)

Commented by mrW1 last updated on 30/Jun/17

In princip the answer is the same as

mine . The locus of M is KL which is

parallel to {XY}^{'} (but the figure shows

KL / / XY, this is wrong)

And my answer is much more exact,

it tells you the different situations for

the locus of M depending on the ranges for

k : k_{1} till t_{4} . My answer tells also how

to determine the position of locus of

M for a certain value of k .

The answer of book mentions only

that the solution could be empty set,

but not when . My answer tells exactly

when, i . e . if k < k_{1} or k > k_{4} and gives

the values of k_{1} and k_{4} as well as other

special k - values k_{2} and k_{3} when the

locus of M is a single point (k = k_{1}, k = k_{4}), part of a

segment from AB to CD (k_{1} < k < k_{2}, k_{3} < k < k_{4}) or a whole

segment from AB to CD (k_{2} ⩽ k ⩽ k_{3}) .

But the core of both solutions is the

same : the locus of M is a straight line

parallel to {XY}^{'} (= A^{'} D^{'} by me) with

TX = AB (= {OA}^{'} by me) and

{TY}^{'} = 2 CD (= {OD}^{'} by me)

Certainly my answer treats also the

case if AB / / CD .

Commented by Tinkutara last updated on 30/Jun/17

Thanks Sir! I knew your solution was

correct . But I posted the solution in my

book to show that it has a good

collection of questions but not solutions .

Commented by mrW1 last updated on 30/Jun/17

They are really interesting questions .

They are really interesting questions .

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