# The many ways to construct a triangle IV

### Constructions by Jacopo (Jack) D'Aurizio

m

_{a}, m_{b}, cSince the medians cut themselves in a 2/3-ratio, this is a SSS problem.

m

_{a}, m_{b}, bFor the same reason, this is a SSS problem with side lengths

(2m _{a}/3 , b/2 , m_{b}/3).R, h

_{a}, aDraw an isosceles triangle with side lengths (R,R,a) and vertex O. Cut the circumcircle with a line parallel to a at distance h

_{a}.R, m

_{a}, aDraw an isosceles triangle with side lengths

(R, R, a) and vertex O. Draw a circle with center in the midpoint of a and radius m_{a}and intersect it with the circumcircle.A, a, m

_{a}Draw the circle that sees a under an angle equal to A. Intersect it with a circle centered in the midpoint of a having radius m

_{a}.A, B, h

_{c}Draw the altitude h

_{c}and a line l perpendicular to it. Draw the circles that see h_{c}under angles equal to B,C and intersect them with l.A, h

_{a}, l_{a}Draw a right triangle having side lengths h

_{a}, l_{a},sqrt(l² Intersect the last side with the lines forming angles equal to +A/2 ,-A/2 with l_{a}- h²_{a})._{a}.A, a, r

The distance between I and a is r, the angle ∠BIC is (pi+A)/2, so we can solve the problem by intersecting a circle and a line parallel to a.

m

_{a}, h_{a}, h_{b}We draw a circle having m

_{a}as diameter, then intersect it with two circles centered in the endpoints of m_{a}and having radii equal to h_{a}, h_{b}/2.In this way we determine the feet h

_{a}of the altitude h_{a}and the direction of the side b, so we find the vertex C, then the vertex B.m

_{a}, h_{b}, h_{c}We draw two opposite right triangles sharing a common hypotenuse m

_{a}and having sides opposite to vertex A with lengths h_{b}/2, h_{c}/2. In this way we determine the lines containing the sides(b, c), having only to find two points on these lines with the proper distances(h from_{c}, h_{b})(b, c). m

_{a}, h_{a}, m_{b}We draw a right triangle having side lengths

(h We call l the line containing the last side, and G the point on m_{a}, m_{a}, sqrt(m_{a}² - h_{a}²))._{a}such that 3 AG = m_{a}. We intersect a circle centered in G with radius 2m_{b}/3 with the line l, finding B, then C.m

_{a}, m_{b}, h_{c}We glue two right triangles such that the triangle GAB satisfies

- GA = 2m
_{a}/3 - GB = 2m
_{b}/3 - d(G, AB) = h
_{c}/3

then call (U, V) the midpoints of

(GA, GB). We call m_{a}the symmetric of U with respect to G, and m_{b}the symmetric of V with respect to G. Am_{b}and Bm_{a}concur in C.- GA = 2m
a, h

_{b}, RWe draw an isosceles triangle with side lengths

(R, R, a) and vertex O. We draw a circle having a as a diameter, and find a point h_{b}on it such thatBH The intersection of the line CH_{b}= h_{b}._{b}with the circumcircle is A.h

_{a}, l_{a}, bWe build a triangle ACH

_{a}havingAC = b, AH CH_{a}= h_{a},_{a}perpendicular to AH_{a}. On the line CH_{a}we determine l_{a}such thatAL We call D the symmetric of C with respect to the line Al_{A}= l_{a}._{a}. The intersection of AD and CL_{A}is B.A, h

_{a}, h_{b}We build a right triangle ABH

_{b}havingAH _{b}= h_{b},∠BAH AH_{b}= A,_{b}perpendicular to BH_{b}. We find a point h_{a}such that BH_{a}is perpendicular to AH_{a}andAH by drawing the circle having AB as a diameter. The intersection of BH_{a}= h_{a}_{a}and AH_{b}is C.m

_{a}, h_{a}, l_{a}We draw M

_{a}, H_{a}and L_{a}such that they lie on a line l perpendicular to AH_{a}and satisfy AH_{a}= h_{a},AM _{a}= m_{a},AL We intersect the perpendicular to l through M_{a}= l_{a}._{a}(the perpendicular bisector of BC) with the line AL_{a}, finding E, that is the midpoint of the arc BC in the circumcircle (little folklore apart: in the Italy IMO team this fact is generally called "Tiozzo's lemma", from the great contestant Giulio Tiozzo). The intersection of the perpendicular bisector of AE with the line EM_{a}is O, the circumcenter. So we draw the circumcircle with radius OA and find B and C on l.

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