The many ways to construct a triangle IV
Constructions by Jacopo (Jack) D'Aurizio
ma, mb, c
Since the medians cut themselves in a 2/3-ratio, this is a SSS problem.
ma, mb, b
For the same reason, this is a SSS problem with side lengths
(2ma/3 , b/2 , mb/3).
R, ha, a
Draw an isosceles triangle with side lengths (R,R,a) and vertex O. Cut the circumcircle with a line parallel to a at distance ha.
R, ma, a
Draw an isosceles triangle with side lengths
(R, R, a)and vertex O. Draw a circle with center in the midpoint of a and radius ma and intersect it with the circumcircle.
A, a, ma
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 ma.
A, B, hc
Draw the altitude hc and a line l perpendicular to it. Draw the circles that see hc under angles equal to B,C and intersect them with l.
A, ha, la
Draw a right triangle having side lengths ha, la,
sqrt(l²a - h²a).Intersect the last side with the lines forming angles equal to +A/2 ,-A/2 with la.
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.
ma, ha, hb
We draw a circle having ma as diameter, then intersect it with two circles centered in the endpoints of ma and having radii equal to ha, hb/2.
In this way we determine the feet ha of the altitude ha and the direction of the side b, so we find the vertex C, then the vertex B.
ma, hb, hc
We draw two opposite right triangles sharing a common hypotenuse ma and having sides opposite to vertex A with lengths hb/2, hc/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 (hc, hb)from (b, c).
ma, ha, mb
We draw a right triangle having side lengths
(ha, ma, sqrt(ma² - ha²)).We call l the line containing the last side, and G the point on ma such that 3 AG = ma. We intersect a circle centered in G with radius 2mb/3 with the line l, finding B, then C.
ma, mb, hc
We glue two right triangles such that the triangle GAB satisfies
- GA = 2ma/3
- GB = 2mb/3
- d(G, AB) = hc/3
then call (U, V) the midpoints of
(GA, GB).We call ma the symmetric of U with respect to G, and mb the symmetric of V with respect to G. Amb and Bma concur in C.
a, hb, R
We 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 hb on it such that BHb = hb.The intersection of the line CHb with the circumcircle is A.
ha, la, b
We build a triangle ACHa having
AC = b, AHa = ha,CHa perpendicular to AHa. On the line CHa we determine la such that ALA = la.We call D the symmetric of C with respect to the line Ala. The intersection of AD and CLA is B.
A, ha, hb
We build a right triangle ABHb having
AHb = hb, ∠BAHb = A,AHb perpendicular to BHb. We find a point ha such that BHa is perpendicular to AHa and AHa = haby drawing the circle having AB as a diameter. The intersection of BHa and AHb is C.
ma, ha, la
We draw Ma, Ha and La such that they lie on a line l perpendicular to AHa and satisfy AHa = ha,
AMa = ma, ALa = la.We intersect the perpendicular to l through Ma (the perpendicular bisector of BC) with the line ALa, 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 EMa is O, the circumcenter. So we draw the circumcircle with radius OA and find B and C on l.