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Let $\overrightarrow {OA}$ and $\overrightarrow {OB}$ be two ray with common end point $O$.
Let $G$ be a point lying in the interior of the $\angle AOB$.
Construct a $\triangle OCD$ such that the sides $OC$ and $OD$ lie on rays $\overrightarrow {OA}$ and $\overrightarrow {OB}$ respectively.

My approach:
$1$$\rightarrow$ Construct $OP$ which passes through $G$ such that $OG:GP=2:1$.
Now taking $P$ as center cut arcs on both rays $\overrightarrow {OA}$ and $\overrightarrow {OB}$ with point of intersection being $M,N$ respectively.
$2$$\rightarrow$ Now draw a line segment $CD\parallel MN$ and passing through $P$ with both $C,D$ lying on rays $\overrightarrow {OA}$ and $\overrightarrow {OB}$.

Hence, $OCD$ is the required triangle.

Proof: Since $CD\parallel MN$ $\Rightarrow$$P$ is midpoint of $CD$ as $MN$ is chord of the circle with $P$ as center.

Question: This question came in one of our regional maths exams for $17$ marks but I got $3$ marks for this proof .
Am I wrong with the construction/proof anywhere? I want to confirm.

Love Invariants
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  • You draw a circle with center $P$ but, what is its radius? Also, how is it sure that it cuts $\vec{OA}$ and $\vec{OB}$? – Vassilis Markos May 18 '18 at 19:15
  • @ΒασίληςΜάρκος- It is a circle with any radius less than OP – Love Invariants May 18 '18 at 19:20
  • But, if you let arbitrarily small radiuses, it is evident that there are cases were the circle does not cut neither $\vec{OA}$ nor $\vec{OB}$. And, after that, how do you choose $C,D$. I mean, they are supposed to be unique, so you cannot just draw any segment parallel to $MN$. – Vassilis Markos May 18 '18 at 19:25
  • @ΒασίληςΜάρκος Btw I wrote it should cut the rays and by triangle inequality there always exists such a radius – Love Invariants May 18 '18 at 19:29
  • I still insist that $C,D$ are not uniquely defined in this way, which, for sure, cannot lead to a right solution, since, given the data of the question, $C,D$ are unique. – Vassilis Markos May 18 '18 at 19:31
  • C,D are lying on the line parallel to MN and passing through P – Love Invariants May 18 '18 at 19:32
  • I think you should add all these details to your question, since, especially the last one, is not clear as it is writen right now. – Vassilis Markos May 18 '18 at 19:33
  • Oops. I forgot to add this. Now its edited – Love Invariants May 18 '18 at 19:41
  • Also, from which theorem does you claim in your proof come from, since I cannot remember of such a result ($CD$ is not a chord of the circle). – Vassilis Markos May 18 '18 at 19:51
  • MN is chord of circle. Draw a perpendicular from P to MN. Now that perpendicular will bisect MN but CD is parallel to MN so that perpendicular will bisect CD too. – Love Invariants May 18 '18 at 20:03
  • Why is that? I mean, try to make this construction with $G$ arbitrarily close to one ray, let's say $\vec{OA}$. Then, the angle $POA$ is smaller than $POB$, so, since $OP$ is not neccessarily perpendicular to $MN$, we have no such result. Moreover, there are $4$ pairs of possible $M,N$, in the general case, and they do have different slopes, so, again, it is not clear, which pair you want. Try to draw some extreme cases to see what I'm saying. – Vassilis Markos May 18 '18 at 20:22
  • @ΒασίληςΜάρκος Where OP is perpendicular to MN is written? – Love Invariants May 18 '18 at 20:28
  • That perpendicular could be named as PQ or something like that. It is obviously not OP. – Love Invariants May 18 '18 at 20:35
  • That's what I'm talking about, it's not perpendicular, so since the angles $POA$ is smaller than $POB$ in the above case, $PC$ will be smaller than $PD$, so $P$ is not the middle of $CD$. Again, draw some cases similar to this to see what I'm trying to imply and consult @dan_fulea's answer for a proper construction. – Vassilis Markos May 19 '18 at 05:13

1 Answers1

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The 3 marks are for finding $P$. Now consider parallels through $P$ to the rays. They intersect the rays in two points, $M,N$. ($OMPN$ is a parallelogram.) Now build $C,D$ on the rays, so that $M,N$ are middle points for $OC$, $OD$.

($MN$ is thus mid line in $OCD$, $CD\| MN$, $P$ is in the middle of $CD$. Use the parallelogram $OMPN$ to show this.)

dan_fulea
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  • My question is what is wrong with my proof? – Love Invariants May 18 '18 at 19:22
  • The question should also specify that the $G$ should be the centroid of the triangle $OCD$ to be constructed. The solution above also constructs two points $M,N$, in a different way. I shall denote them here in the comment by $M^, N^$, these are my points. In both cases, we consider after this prelude the parallel through $P$ to $M^N^$, respectively $MN$. So this is the difference, and this points to the error. Just draw a figure starting from a parallelogram, even better, a rectangle on maths paper, $OM^PN^$ with $OM^$ much bigger than $ON^$. Now try to make your construction. – dan_fulea May 18 '18 at 19:53
  • In this condition arcs will be cut outside the triangle OCD. – Love Invariants May 18 '18 at 20:12
  • We still have no idea what are the "arcs". For instance, let explicitly the rays be the axes $Ox$, $Oy$. Let us fix ideas and the picture with the point $P$ with coordinates $(12,5)$. It has distance $13$ to the origin. I hope it is clear that $C,D$ are the points $C(24,0)$, and $D(0,10)$. Just draw all points, best also the point $X(24,10)$, and the four rectangles inside the big rectangle $OCXD$. Now look at your construction. What do we get for "arcs" of radius $4$? Nothing. This is your problem, should be eliminated. Radius $5,6,7,11$? Nothing. Radius $12$? Two intersections with $Ox$... – dan_fulea May 18 '18 at 20:44
  • Oh that is what you mean. I wrote in my solution that arcs can be cut on the opposite side of O. Just extend OA,OB and cut an arc with radius more than OP. It will definitely cut OA,OB. – Love Invariants May 18 '18 at 21:03
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    Yes, but this "random cut" does not lead to the (unique) solution. Moreover, different random choices for the radius of the circle centered in $P$ lead to different solutions $C,D$, this cannot be true. Please use some geometry software like geogebra to see how the figure changes completely. – dan_fulea May 18 '18 at 23:06
  • Again, adding to the last @dan_fulea's comment, we usually have four intersection points, even if it was a right construction, you would need to uniquely define which of all these you are talking about. – Vassilis Markos May 19 '18 at 05:16
  • @ΒασίληςΜάρκος-Out of 4 intersection points 2 will be on the triangle and 2 outside. Both taken together will give same result. Also I will try it using geogebra. – Love Invariants May 19 '18 at 17:42