Jack has a quadrilateral that consists of four sticks. It turned out that Jack can form three different triangles from those sticks. Prove that he can form a fourth triangle that is different from the others.
Problem
Source:
Tags: geometry, Sides of a triangle
IAmTheHazard
21.06.2020 19:57
I don't really get the problem. He makes a quadrangle with 4 sticks (I got that). Then he picks 3 of the 4 sticks, makes a triangle, and repeats twice until he has made 3 triangles in total. Then the problem asks if he can repeat this a third time (for 4 triangles total)?
parmenides51
21.06.2020 20:02
What I understand is that with sidelenghts $a,b,c,d$ one may create triangles using $a-b-c$ or $a-b-d$ or $a-c-d$. It asks whether it is possible to create a triangle using $b-c-d$, such as all triangles are different each time.
parmenides51
21.06.2020 20:19
trapezoid $ABCD$, $BC\parallel AD$, $\angle A=\angle B=90^o$, $AB=4, BC=3, CD=5, DA=6$
all 4 triangles exist because longest less than sum of other two sides,
a.k.a. $5<3+4, 6<3+4,6<5+4, 6<3+5$
.
parmenides51
22.06.2020 23:08
all our above comments were about my wrong translation parmenides51 wrote: Cyril had a quadrangle made up of four sticks. He dismantled it and, choosing three sticks, made three different triangles. Is it true that he can make a fourth triangle, different from the ones made earlier? IAmTheHazard wrote: I don't really get the problem. He makes a quadrangle with 4 sticks (I got that). Then he picks 3 of the 4 sticks, makes a triangle, and repeats twice until he has made 3 triangles in total. Then the problem asks if he can repeat this a third time (for 4 triangles total)? parmenides51 wrote: What I understand is that with sidelenghts $a,b,c,d$ one may create triangles using $a-b-c$ or $a-b-d$ or $a-c-d$. It asks whether it is possible to create a triangle using $b-c-d$, such as all triangles are different each time. parmenides51 wrote:
trapezoid $ABCD$, $BC\parallel AD$, $\angle A=\angle B=90^o$, $AB=4, BC=3, CD=5, DA=6$
all 4 triangles exist because longest less than sum of other two sides,
a.k.a. $5<3+4, 6<3+4,6<5+4, 6<3+5$
. I just added the correct translation thanks to JosefSvejk
Grandmaster_Gogo
23.06.2020 00:38
Well what you did wasn't exactly correct . You proved it using just one example ... You should now instead of some integers sides and fixed angles and whatsoever, now you should use just some variable sides such as a,b,c and d. And then prove it. Cause in the question they have just mentioned a quadrilateral , neither a trapezoid,a rectangle or something else of that kind. You may then use angles and so and so , but that would become tedious, so you should rather use inequalities.
parmenides51
23.06.2020 06:24
my reply, responds to the first incorrent translation mentioned in #5. as a reply to the current correct translation, obviously it is wrong, not even close to a solution