Difference between revisions of "1991 OIM Problems/Problem 6"

Revision as of 18:44, 22 December 2023

Problem

Given 3 non-aligned points $M$ , $N$ and $H$ , we know that $M$ and $N$ are midpoints of two sides of a triangle and that $H$ is the point of intersection of the heights of said triangle. Build the triangle.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

Case 1: $\angle NHM = 90^{\circ}$

If you measure $\angle NHM$ on the given points and it happens to be a right angle, then constructing the triangle is easy because point $H$ is also point $A$ of the triangle $ABC$ . One can notice if this angle is a right angle or not if you can draw a perpendicular from point $M$ to line $NH$ and it passes through $H$ . If this happens to be the case, then since $AM=MB$ and $AN=NC$ then one can simply draw a circle with the compass at points $M$ and $N$ with radiuses measuring $MA$ and $NA$ respectively. Then extend the lines $AM$ and $AN$ to the intersection on their respective circles at $B$ and $C$ respectively. Then draw triangle $ABC$ .

Case 2: $\angle NHM \ne 90^{\circ}$

Let the red circle in the image above be the circumcircle of triangle $ABC$ . Let $BD$ be a diameter of the circle. This means that $\angle BAD$ and $\angle BCD$ are both equal to $90^{\circ}$ because right angle triangles inscribed in circles with the hypothenuse on the diameter. Therefore $AH$ is parallel to $DC$ and $CH$ is parallel to $CD$ . Thus quadrilateral $ADCH$ is a parallelogram with $N$ in the center and $HN=ND$ . So, one can draw point $D$ using $N$ and $H$ . Since $\angle MAD=90^{\circ}$ , then $MD$ is the diameter of a circle that also passes through $A$ . This means that one can find point $A$ from the intersection of this circle and the perpendicular to $MN$ that passes through $A$ and we can now start our construction as follows:

Given the points $M$ , $N$ , and $H$ , we first draw a perpendicular from point $H$ to $MN$ with straight edge and compass using the traditional known method of finding tow points from $H$ on $MN$ equidistant to $H$ and from those two points drawing the perpendicular bisector to them. That way we can draw the black perpendicular line.

Then, with the compass draw the red circle with center at $N$ and radius $NH$ . From $H$ we extend the line $HN$ and where it intersects the red circle that's our point $D$ . We then draw blue line $MD$ and find it's bisection point using traditional bisection method with compass and straight edge. From this bisection point and the compass using that as its center we can draw the blue circle. This blue circle intersects the black perpendicular line at $A$ . Since $AM=MB$ and $AN=NC$ then one can simply draw a circle with the compass at points $M$ and $N$ with radiuses measuring $MA$ and $NA$ respectively. Then extend the lines $AM$ and $AN$ to the intersection on their respective circles at $B$ and $C$ respectively. Then draw triangle $ABC$ .

But the given points do not give us one unique triangle. It gives us two because the blue circle also intersects the black perpendicular line at another point. We will call this point $A'$ . Then using the same methods as described in the procedure for triangle $ABC$ we find triangle $A'B'C'$ which also has the given points $M, N$ and $H$

And this provides all of the cases and all of the triangles we can draw from those given points.

Note. I actually competed at this event in Argentina when I was in High School representing Puerto Rico. If I remember correctly I solved the case 1 described here as it was the easiest and then wasted a lot of time trying to figure out a way unsuccessfully. I also noted that two triangles could be build instead of just one. I think they awarded me 3 points out of 10.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

@@ Line 15: / Line 15: @@
 '''Case 2:''' <math>\angle NHM \ne 90^{\circ}</math>
-[[File:1991_OIM_P6b.png|500px]]
+[[File:1991_OIM_P6b.png|400px]]
 Let the red circle in the image above be the circumcircle of triangle <math>ABC</math>.  Let <math>BD</math> be a diameter of the circle.  This means that <math>\angle BAD</math> and <math>\angle BCD</math> are both equal to <math>90^{\circ}</math> because right angle triangles inscribed in circles with the hypothenuse on the diameter.  Therefore <math>AH</math> is parallel to <math>DC</math> and <math>CH</math> is parallel to <math>CD</math>.  Thus quadrilateral <math>ADCH</math> is a parallelogram with <math>N</math> in the center and <math>HN=ND</math>. So, one can draw point <math>D</math> using <math>N</math> and <math>H</math>.  Since <math>\angle MAD=90^{\circ}</math>, then <math>MD</math> is the diameter of a circle that also passes through <math>A</math>.  This means that one can find point <math>A</math> from the intersection of this circle and the perpendicular to <math>MN</math> that passes through <math>A</math> and we can now start our construction as follows:
+[[File:1991_OIM_P6c.png|400px]]
+Given the points <math>M</math>, <math>N</math>, and <math>H</math>, we first draw a perpendicular from point <math>H</math> to <math>MN</math> with straight edge and compass using the traditional known method of finding tow points from <math>H</math> on <math>MN</math> equidistant to <math>H</math> and from those two points drawing the perpendicular bisector to them.  That way we can draw the black perpendicular line.
+Then, with the compass draw the red circle with center at <math>N</math> and radius <math>NH</math>.  From <math>H</math> we extend the line <math>HN</math> and where it intersects the red circle that's our point <math>D</math>.  We then draw blue line <math>MD</math> and find it's bisection point using traditional bisection method with compass and straight edge.  From this bisection point and the compass using that as its center we can draw the blue circle.  This blue circle intersects the black perpendicular line at <math>A</math>.  Since <math>AM=MB</math> and <math>AN=NC</math> then one can simply draw a circle with the compass at points <math>M</math> and <math>N</math> with radiuses measuring <math>MA</math> and <math>NA</math> respectively.  Then extend the lines <math>AM</math> and <math>AN</math> to the intersection on their respective circles at <math>B</math> and <math>C</math> respectively.  Then draw triangle <math>ABC</math>.
-* Note.  I actually competed at this event in Argentina when I was in High School representing Puerto Rico.  I think I may have been able to build some cases of this.  I don't remember much of it.
+[[File:1991_OIM_P6d.png|400px]]
+But the given points do not give us one unique triangle.  It gives us two because the blue circle also intersects the black perpendicular line at another point.  We will call this point <math>A'</math>.  Then using the same methods as described in the procedure for triangle <math>ABC</math> we find triangle <math>A'B'C'</math> which also has the given points <math>M, N</math> and <math>H</math>
+And this provides all of the cases and all of the triangles we can draw from those given points.
+* Note.  I actually competed at this event in Argentina when I was in High School representing Puerto Rico.  If I remember correctly I solved the case 1 described here as it was the easiest and then wasted a lot of time trying to figure out a way unsuccessfully.  I also noted that two triangles could be build instead of just one.  I think they awarded me 3 points out of 10.
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Difference between revisions of "1991 OIM Problems/Problem 6"

Revision as of 18:44, 22 December 2023

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