Figure 1 shows an equilateral triangle with side length equal to  $1$. Two squares of side length  $a$  and  $2a$  placed side by side just fit inside the triangle as shown.
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Question: Figure shows an equilateral triangle with side length equal to  $1$ . Two squares of side length  a  and $2a$  placed side by side just fit inside the triangle as shown. Find the exact value of  $a$.  Its an Assessment question from edX course &quot;A-Level Mathematics Course 1&quot;  and I am supposed to use skills that I learnt in Indices and surds,Inequalities and The Factor Theorem. I have tried finding the height of triangle and then use similar triangles to find the right triangle length still No luck. I am just looking for food for thought or very small hints thats all.
100171
Prove that given $a,b,c &gt; 0$, it is possible to construct a triangle with sides of length $a,b,c$ if and only if $pa^2+qb^2 &gt; pqc^2$ for any $p,q$ with $p+q = 1$.   Should I prove this using the triangle inequality? Also should I split the proof up into two parts for the if and only if?
30795
Let $a,b,c$ be the lengths of the sides of a triangle. The area is given by Heron's formula: $$A = \sqrt{p(p-a)(p-b)(p-c)},$$ where $p$ is half the perimeter, or $p=\frac{a+b+c}{2}$.  Could you please provide the proof of this formula? Thank you in advance.
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I want to calculate the area of the largest square which can be inscribed in a triangle of sides $a, b, c$ . The "square" which I will refer to, from now on, has all its four vertices on the sides of the triangle, and of course is completely inscribed within the triangle, and by "largest square", with the maximum area of such squares.   I did some research. I found that in the case of right triangles, there seemed to be only two kinds of squares being formed. One with a side of $\frac{ab}{a+b}$ and the other of $\frac{abc}{c^2 + ab}$, $c$ being the hypotenuse.   After some time, I realized that there was a common formula for all triangles. And that is:  $$side = \frac{ch}{c+h}$$       or $$\frac{1}{side} = \frac{1}{c} + \frac{1}{h}$$ where, $h$ is height on base $c$, or $$h = \frac{2\sqrt{(s(s-a)(s-b)(s-c))}}{c}$$    I achieved the above results using similar triangles, and have added a figure, to make the point clear. But that is all, I can prove. I tested for many triangles, and found that the smaller side always yields the larger square, but I am not able to prove it. For example, I can say that the largest square of a $4,7,10$ triangle, has area of approximately $\fraq{16}{3}$. I also want to make another conjecture (with an intuitive proof) that any triangle will give at most three squares.  So I want to ask you, are my calculations, proofs ... correct? Have I got them right? And those predictions? Are they right, and if so, proofs? Conclusively, do we have a sure-fire formula to get the area of the largest square?
115758
The problem statement is as follows:  A triangle is dissected into six smaller triangles by its angle bisectors. Prove that the intersections of the angle bisectors of each of these smaller triangles lie on an ellipse.  Could anybody here solve it?
115265
Problem:    How many triangles do $m$ lines form if   a) Every two lines intersect and no three lines intersect at one point.  b) There are $n$ lines among $m$ lines that are parallel to each other. No other line is parallel to these $n$ lines, and no other two lines are parallel to each other. Again no three lines intersect at one point.  Thank you.
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In the quadrangle $ABCD$, the points $E,F,G,H$ are the midpoints  of respectively $AB, BC, CD, DA$. We know that area $\triangle AHL=a$, $\triangle DIG=b$, $\triangle FJC=c$, $\triangle EBK=d$. Prove that area $IJKL=a+b+c+d$.
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In grade school we learn that a square can be equidecomposed into two congruent right isosceles triangles. Does the following three dimensional generalization hold?  Consider a trirectangular tetrahedron with vertices at $(0,0,0)$, $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ and volume $\frac{1}{6}$. (It has three orthogonal right triangular faces and one equilateral triangular face opposite the (tri-right-angled) common vertex.) The inscribing cube of the tetrahedron has unit side length and unit volume. Is there a dissection of this cube into 6 tetrahedra of the above form?   Provided that such a dissection exists and this is the way it is to be done, it's easy to see how 4 tetrahedra can be glued together to give the boundary of the cube, but I don't quite see how the other two should fit inside.
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Let $a,b,c$ be the sides of a triangle. Show that: $$(\sqrt a + \sqrt b - \sqrt c)(\sqrt a - \sqrt b + \sqrt c)(-\sqrt a + \sqrt b + \sqrt c) \ge \sqrt {(a+b-c)(a-b+c)(-a+b+c)}.$$
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Someone asked me this question which I am really stuck at, any help is appreciated.     If $a,b,c,d$ are the sides of a  quadrilateral and $m,n$ are diagonals of the quadrilateral, then prove that area of the quadrilateral is $$\frac14[4m^2n^2-(b^2+d^2-a^2-c^2)^2]^{\frac12}$$   I do not recall anything like this from my school days. I need some hints to get started on this, I do not remember high school geometry formulas, may be one of them could help here.  It works fine for squares and recatngles, so I assume it is a valid formula.
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Triangle $ABC$ is an equilateral triangle and $P$ is an arbitrary point inside it. The distance from $P$ to $A$ is $4$ and the distance from $P$ to $B$ is $6$ and the distance from $P$ to $C$ is $5$. How to find the side of an equilateral triangle from this information?
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Show that $\lvert e^{i\theta} - 1\rvert = 2\lvert\sin(\theta/2)\rvert$ by using the geometry of the triangle with vertices 0, 1, and the midpoint of the line joining 0 and $e^{i\theta}$.  I have been able to show this identity through other means, however I am stuck on how to utilize this particular triangle.
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I was looking at * and started thinking about the case of a polygon with 4 sides: a square.   The question asks for a program that can take a polygon of N sides and return the smallest possible polygon of N-1 sides that contains the original polygon.  Thinking about it, I wasn't sure what the optimal triangle encapsulating a square would look like. In KSab's answers, he returns a triangle that looks like this:    This area is 2*(area of the square). With some easy guess-and-check work I found that this seems to be optimized. How do you prove that there is no better solution?  EDIT: another consideration for the orientation of the triangle could be this:
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In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is $(A)$equilateral $(B)$isosceles $(C)$right angled $(D)$none of these    The given condition is $a^2+b^2+c^2=ac+ab\sqrt3$. Using sine rule, $a=2R\sin A,b=2R\sin B,c=2R\sin C$,we get $\sin^2A+\sin^2B+\sin^2C=\sin A\sin C+\sin A\sin B\sqrt3$  I am stuck here.
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"Given any point on a ellipse, is it always possible to inscribe an equilateral triangle, with a vertex coincident with that point, in the ellipse?"  I thought I could use analytical geometry, but when I first come up with a line-long equation I gave it up. Do you think there are faster, and more elegant ways to solve this problem?
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Question: If $x, y, z$ are the side lengths of a triangle, prove that $x^2 + y^2 + z^2 &lt; 2(xy + yz + xz)$        My solution: Consider  $$x^2 + y^2 + z^2 &lt; 2(xy + yz + xz)$$  Notice that $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)$  Hence  $$(x+y+z)^2-2(xy+yz+xz) &lt; 2(xy + yz + xz)$$  $$ (x+y+z)^2 - 4(xy+yz+xz) &lt; 0 $$  As $x,y,z$ are all greater than zero as a side length of a triangle can not be negative. And because $(x+y+z)^2 &gt;0$ for all real $x,y,z$ therefore the whole expression is less than zero. $Q.E.D$  Am I correct? Or could I be more "rigorous" I am a highschool student and getting into proof so any tips would be appreciated as well :)
9752
Generally,the length of the sum of two vectors is not equal to the sum of their lengths. To see this consider the vectors $u$ and $v$ as shown below.  By considering $u$ and $v$ as two sides of a triangle, we can see that the lengths of the third side is $\| u + v \|$ and we have $\| u + v \| \leq \|u\| + \|v\|$.  Under what circumstance equality occurs and how can one prove that?