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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

	$They are really interesting questions .$
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