Tag Info

Hot answers tagged

5

We use the well-known formula $R = \frac{abc}{4A}$, where $A$ is the area of the triangle. Thus $8R^2 = \frac{8a^2b^2c^2}{16A^2}$, and by Heron's formula $$16A^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c).$$ Now assume that $a^2 + b^2 + c^2 = 8R^2$. Then we have $$8a^2b^2c^2 - (a^2 + b^2 + c^2)(a+b+c)(-a+b+c)(a-b+c)(a+b-c) = 0.$$ Expanding this gives $$a^6 + b^6 + c^6 ...


5

I decided to upgrade my comment into an answer. Angle $C$ is free to vary. That would change the length of $BD$ while preserving the given lengths, which is were you went astray (the length of $BD$ is not necessarily $4$). Now, in general, if we know two side lengths of a triangle (call them $x$ and $y$) and the angle measure between them (call it ...


3

In your case, the area of the triangle can be made to be greater than $72$ if the two remaining sides both have length $14.$ That is, if the triangle is $12-14-14,$ then it is isosceles, so we can find its area pretty easily: Drop an altitude to the $12$-length side, dividing that side into two segments of length $6.$ Then the altitude has length ...


3

Assuming that $R$ means circumradius we can use the formula $$R=\sqrt{\frac{a^2+b^2+c^2}{8(1+\cos A\cos B\cos C)}},$$ where $a$, $b$, and $c$ are the sidelengths and $A$, $B$, and $C$ are the angles. Plugging in $a^2+b^2+c^2=8R^2$, we reduce the above to $$1+\cos A\cos B\cos C=1,$$ whence one of $\cos A$, $\cos B$, or $\cos C$ must be $0$ - meaning that ...


3

Embed the construction in the complex plane. Let $\omega=\exp\left(\frac{\pi i}{3}\right),B=0,C=1,E=1+v$. Then $A=\omega$ and $D=1+\omega v$, hence $F=B+E-A$ implies: $$ F = 1-\omega+v,$$ hence: $$ \omega F = \omega -\omega^2 + \omega v = 1+\omega v = D, $$ so $BFD$ is equilateral.


3

From the Wikipedia article on Median, the lengths of the medians in terms of the sidelengths are: $m_a = \dfrac{1}{2}\sqrt{2b^2+2c^2-a^2}$ $m_b = \dfrac{1}{2}\sqrt{2c^2+2a^2-b^2}$ $m_c = \dfrac{1}{2}\sqrt{2a^2+2b^2-c^2}$ Solving for $(m_a,m_b,m_c)$ in terms of $(a,b,c)$ yields: $a = \dfrac{2}{3}\sqrt{2m_b^2+2m_c^2-m_a^2}$ $b = ...


2

The proof is ugly due to the limitation of the Ecludean framework. I try to break it down into 3 diagrams. See the first. As mentioned, $\triangle AEC$ is congruent to $\triangle BDC$. [Proof is therefore skipped.] Then $\alpha = \beta$ [result #1] And BF = AE = BD [result #2] See figure 2. Join EF. Through O, draw MN // AE cutting EF at N. HOK is ...


2

Hint: What is the area of the circle? If the line $BC$ passes through the center of the circle, what area is on either half of the line? And finally, what is the area of the triangle?


2

In the disc model take two lines $L_1,L_2$ emanating from the center $0$, and forming an angle $\alpha$ berween them. Now, consider the point $x_t\in L_1$ ad distance $t$ from $0$ and the line $L_3(t)$ emanating from $x_t$ and forming an angle $\beta$ with $L_1$. Let $y_t=L_3(t)\cap L_2$. Now, $y_t$ can be a single point or the empty set. If $y_t$ is a ...


2

Let $S$ be the area of the triangle. Since we have $4RS=abc$ and $$\begin{align}16S^2&=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)\\&=-(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+c^2a^2)\end{align}$$ if $a^2+b^2+c^2=8R^2$, then we have $$\begin{align} 0 ...


2

It should be $\cos B=\frac 47$


2

Yes, there is a way. Due to the Euler theorem, $O,G,H$ are collinear and $HO=3\,GO$. This implies that if we take $O$ as the origin, the vector equation: $$ H = A+B+C $$ holds. By applying the triangular inequality: $$ OH = |H| \leq |A|+|B|+|C| = OA+OB+OC = 3R.$$


2

From this, $\displaystyle IO^2=2r^2+4R^2-S_w,$ where $\displaystyle2S_w=a^2+b^2+c^2$ From this, $\displaystyle SO^2=9R^2-2S_w$ So, $\displaystyle SO^2-2IO^2=R^2-4r^2$ But from this, $\displaystyle R\ge 2r$


2

$a^2+b^2=5^2$ has many solutions, $b=\sqrt{25-a^2}$ for any $a\le 5$.


2

Consider circumscribed circle and it's radius $R$. By inscribed angle theorem you get, that $|c|=|R|$, where $c$ is third side of your triangle $a=b=10$. Now you have formula $\displaystyle S=\frac{abc}{4R}$, where $S$ is area of triange. So: $$S=\frac{10 \cdot 10 \cdot c}{4R}=\frac{100}{4}=25$$


2

Use Pythagoras: $$ a^2 + b^2 = c^2 $$ Pythagorean theorem $$a^2+12^2=20^2$$ $$a^2 + 144 = 400$$ Subtract 144 from both sides. $$a^2 = 256$$ you only take the positive answer. $$a = \sqrt{256}=16$$


1

The line has to be drawn just through the midpoint of $BC$. A median splits a triangle in two triangles having the same base and the same height, hence equal areas. To find the length of a median you can use the formula: $$ m_a^2 = \frac{2b^2+2c^2-a^2}{4},$$ giving in your case: $$ m_a = \frac{1}{2}\sqrt{2\cdot 3.9^2+2\cdot 3.2^2-5.7^2}=2.145\ldots$$


1

This problem gave me a really hard time, lots of steps are involved. Step 1. If $ABC$ is equilateral and $MAF,MBD,MCE$ have the same area, then $M$ is the center of $ABC$. Given that $[x,y,z]$ are the trilinear coordinates of $M$, the areas of our triangles are proportional to: $$\frac{xy}{x+z},\frac{yz}{x+y},\frac{xz}{y+z}$$ hence $M=[1,1,1]$ is the ...


1

Try drawing an altitude from one of the $75$ degree angles to the opposite 10-unit side. You should be able to determine the length of this altitude using a 30-60-90 triangle.


1

Since the circumcircle of $ABC$ is the incircle of $A'B'C'$, $O$ is the incenter of $A'B'C'$, hence $A'B',B'C'$ and $C'A'$ are the exterior angle bisectors of $ABC$, and $A'B'C'$ is the excentral triangle of $ABC$. Due to relations $(3)$ and $(4)$ given here: $$ r_1+r_2+r_3 = 4R+r. $$ Proof: Since ...


1

You know that the internal angles of a triangle add up to $180$ and you know $B$ is a right angle so we have $180 = A + B + C = A + 90 + C$ and so $A + C = 90$. We are given that $x+30 = C$ and so $A + x + 30 = 90$ and so $A = 60 - x$.


1

$A$ and $C$ wouldn't be complementary angles in that case. They give it as $x+30^\circ$ because they want you to consider it symbolically (i.e. it would be equally applicable to an angle of $35^\circ$, or $40^\circ$, etc.)


1

The $x$ is there just to make the question slightly abstract. Perhaps the question author thought this bit of abstraction would better test your comfort with triangles and variables. You ask why the answer cannot be d). Let's check by assigning $x$ a particular value, like $0^\circ$. In this case, $C$ has angle $30^\circ$, so we know that $A$ has angle ...


1

Consider the following configuration: in which $H_A,H_B$ are the feet of the altitudes from $A$ and $B$. Since $CH_A HH_B$ is a cyclic quadrilateral we have: $$AH_B\cdot AC = AH\cdot AH_A,\qquad BH_A\cdot BC = BH\cdot BH_B.$$ Moreover: $$BH_A\cot C = HH_A,\qquad AH_B\cot C = HH_B,$$ hence, through circular inversion, we have that our equation implies ...


1

HINT : Do you see why the followings hold? $$\triangle PQR:\triangle MQR=2:1$$ $$\triangle MQR:\triangle SQR=3:2$$ $$\triangle SQR:\triangle MNS=2^2:1^2$$ P.S. For exmaple, since $MQ:SQ=3:2$, we have $\triangle MQR:\triangle SQR=3:2$. Do you see why?


1

Hint 1: $\triangle MNS$ is similar to $\triangle QRS$ (why?) Hint 2: You know the ratio of $MN$ to $QR$. Can you then find the ratio of the areas of $\triangle MNS$ and $\triangle QRS$ Hint 3: $$\frac{\triangle MNS}{\triangle PQR} = \frac{\triangle MNS}{\triangle PNM} \frac{\triangle PNM}{\triangle PQR}$$


1

This is quite an interesting question. Could you share with us which level or type of test or exam this is from? Draw a line from P to S and extend it such that it meets QR at L. Let K be the point of intersection of PL and MN. Note that MN is parallel to QR but half the length (as M and N are midpoints). i.e. $MN=\frac 12 QR$. As such, $PK=\frac 12 ...


1

Converting comments to answer, as requested. The perpendicular from $O$ to $\overline{KL}$ must also be perpendicular to $\overline{MN}$. (Why?) Let $P$ and $Q$ be the points where the perpendicular crosses these segments. Show that $\overline{OP}\cong\overline{OQ}$. (Hint: The perpendicular from $O$ to $\overline{AB}$ meets the segment at its midpoint, ...


1

$\sin B\neq \frac{4}{7}$ $\sin B=\frac{AC}{7}$ $AC=\sqrt{33}$ by pythagorean theorem.


1

Note $\angle AEI=\angle ADI=90^\circ - \frac{\angle A}{2} = \frac{\angle B}{2}+\frac{\angle C}{2}$. It follows that $$\angle DIB=\angle ADI-\angle DBI = \frac{\angle C}{2}=\angle ECI.$$ Thus $\triangle DIB$ and $\triangle ECI$ are similar (Having equal angles). The equality follows.



Only top voted, non community-wiki answers of a minimum length are eligible