Radical Condition Implies Isosceles

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Radical Condition Implies Isosceles

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Source: Black MOP 2012

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#1 h Aug 10, 2023, 4:54 PM • 1 Y

Y by cubres

Prove that any triangle with
$\sqrt{a+h_B}+\sqrt{b+h_C}+\sqrt{c+h_A}=\sqrt{a+h_C}+\sqrt{b+h_A}+\sqrt{c+h_B}$ is isosceles.

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#2 h Aug 10, 2023, 4:56 PM • 1 Y

Y by cubres

Posting for the future visitor.

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This post has been edited 1 time. Last edited by peace09, Aug 10, 2023, 4:57 PM
Reason: wording

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#3 h Aug 10, 2023, 5:08 PM • 1 Y

Y by cubres

my solution from a few months ago

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#4 h Dec 2, 2023, 2:04 PM • 1 Y

Y by cubres

Let $A,B,C$ the three elements on the LHS and $x,y,z$ the elements on the RHS.

First, we can see that

$\begin{align*} A+B+C &=x+y+z\\ AB+BC+CA &=xy+yz+zx\\ ABC&=xyz \end{align*}$
The second equality comes from the fact that :
$(A+B+C)^2-(A^2+B^2+C^2)=(x+y+z)^2-(x^2+y^2+z^2)$ And the last equality comes from a simple expansion. By Vieta, we can deduce that both of the triples are roots of the same polynomial of the third degree. One can check the three cases of equality to prove the desired result.

$\mathbb{Q.E.D.}$

This post has been edited 1 time. Last edited by Cusofay, Dec 2, 2023, 2:04 PM
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#5 h Apr 16, 2024, 10:42 PM • 1 Y

Y by cubres

Solution

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#7 h May 5, 2024, 1:11 PM • 1 Y

Y by cubres

Since the geometry condition is kinda useless we can denote $h_A = \frac{1}{a}$ , $h_B = \frac{1}{b}$ and $h_C = \frac{1}{c}$ (in that way the triangle has area $\frac{1}{2}$ ). Now let $\sqrt {a + \frac{1}{b}} = r_1$ , $\sqrt {b + \frac{1}{c}} = r_2$ and $\sqrt {c + \frac{1}{a}} = r_3$ , and also let $r_1$ , $r_2$ and $r_3$ be the roots of a polynomial P(x). Let $\sqrt {a + \frac{1}{c}} = s_1$ , $\sqrt {b + \frac{1}{a}} = s_2$ and $\sqrt {c + \frac{1}{b}} = s_3$ , and also let $s_1$ , $s_2$ and $s_3$ be the roots of a polynomial Q(x). From the condition of the problem we get that $r_1 + r_2 + r_3 = s_1 + s_2 + s_3$ . Now $r_1r_2r_3 = \sqrt {(a + \frac{1}{b})(b + \frac{1}{c})(c + \frac{1}{a})} = \sqrt{abc + a + b + c + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{abc}}$ . Also $s_1s_2s_3 = \sqrt {(a + \frac{1}{c})(b + \frac{1}{a})(c + \frac{1}{b})} = \sqrt{abc + a + b + c + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{abc}}$ $\Rightarrow$ $r_1r_2r_3 = s_1s_2s_3$ . We have that $r_1^2 + r_2^2 + r_3^2 = a + \frac{1}{b} + b + \frac{1}{c} + c + \frac{1}{a}$ and $s_1^2 + s_2^2 + s_3^2 = a + \frac{1}{c} + b + \frac{1}{a} + c + \frac{1}{b}$ $\Rightarrow$ $r_1^2 + r_2^2 + r_3^2 = s_1^2 + s_2^2 + s_3^2$ . We know that $2(r_1r_2 + r_1r_3 + r_2r_3) = (r_1 + r_2 + r_3)^2 - (r_1^2 + r_2^2 + r_3^2)$ and $2(s_1s_2 + s_1s_3 + s_2s_3) = (s_1 + s_2 + s_3)^2 - (s_1^2 + s_2^2 + s_3^2)$ and since we know $r_1 + r_2 + r_3 = s_1 + s_2 + s_3$ and $r_1^2 + r_2^2 + r_3^2 = s_1^2 + s_2^2 + s_3^2$ , this means $r_1r_2 + r_1r_3 + r_2r_3 = s_1s_2 + s_1s_3 + s_2s_3$ $\Rightarrow$ now we have that $r_1 + r_2 + r_3 = s_1 + s_2 + s_3$ , $r_1r_2 + r_1r_3 + r_2r_3 = s_1s_2 + s_1s_3 + s_2s_3$ , $r_1r_2r_3 = s_1s_2s_3$ , which by Vieta's means that $P(x) \equiv Q(x)$ $\Rightarrow$ the two polynomials have equal roots. Now WLOG $\sqrt {a + \frac{1}{b}} = \sqrt {a + \frac{1}{c}}$ $\Rightarrow$ b = c, WLOG $\sqrt {a + \frac{1}{b}} = \sqrt {b + \frac{1}{a}}$ $\Rightarrow$ a = b, WLOG $\sqrt {a + \frac{1}{b}} = \sqrt {c + \frac{1}{b}}$ $\Rightarrow$ a = c $\Rightarrow$ whatever is the equality in the roots from LHS and RHS there are always two equal sides of the triangle $\Rightarrow$ $\triangle ABC$ is isosceles. We are ready.

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combowomborhombo

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combowomborhombo

#8 h Aug 8, 2024, 10:35 PM • 1 Y

Y by cubres

Our first step is to reduce the number of variables in the problem, specifically to establish a relationship between $h_A$ , $h_B$ , $h_C$ and $a$ , $b$ , $c$ . The geometric condition doesn't impose any specific constraints, so we can assume that the area of the triangle is $\frac{1}{2}$ . Using the area formula, we obtain the relationships:

$\begin{align*} h_A &= \frac{1}{a}, \\ h_B &= \frac{1}{b}, \\ h_C &= \frac{1}{c}. \end{align*}$
Next, consider the two sides of the equation as two separate polynomials. Let the LHS be $P(x)$ , with roots:

$\begin{align*} m_1 &= \sqrt{a + \frac{1}{b}}, \\ m_2 &= \sqrt{b + \frac{1}{c}}, \\ m_3 &= \sqrt{c + \frac{1}{a}}. \end{align*}$
The RHS will be $Q(x)$ , with roots:

$\begin{align*} n_1 &= \sqrt{a + \frac{1}{c}}, \\ n_2 &= \sqrt{b + \frac{1}{a}}, \\ n_3 &= \sqrt{c + \frac{1}{b}}. \end{align*}$
Let's examine the relationships between the roots. The first relationship is:

$\begin{align*} m_1 + m_2 + m_3 &= n_1 + n_2 + n_3 \end{align*}$
This follows from the conditions given in the problem. Another important relationship is:

$\begin{align*} m_1 m_2 m_3 &= n_1 n_2 n_3 \end{align*}$
This can be seen by writing it out explicitly:

$\begin{align*} m_1 m_2 m_3 &= \sqrt{(a + \frac{1}{b})(b + \frac{1}{c})(c + \frac{1}{a})} \\ &= \sqrt{abc + a + b + c + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{abc}} \end{align*}$
and

$\begin{align*} n_1 n_2 n_3 &= \sqrt{(a + \frac{1}{c})(b + \frac{1}{a})(c + \frac{1}{b})} \\ &= \sqrt{abc + a + b + c + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{abc}} \end{align*}$
This shows that $m_1 m_2 m_3 = n_1 n_2 n_3$ .

Our goal is to show that two of $a$ , $b$ , $c$ are equal. We now focus on the pairwise relationships in the polynomial. We know:

$\begin{align*} m_1^2 + m_2^2 + m_3^2 &= n_1^2 + n_2^2 + n_3^2 \end{align*}$
This is true by the commutative property after removing the square roots. From Newton's sums, we have:

$\begin{align*} 2(r_1r_2 + r_1r_3 + r_2r_3) &= (r_1 + r_2 + r_3)^2 - (r_1^2 + r_2^2 + r_3^2) \end{align*}$
Replacing $r_i$ with $m_i$ and $n_i$ , and using our known equalities, we get:

$\begin{align*} m_1 m_2 + m_2 m_3 + m_3 m_1 &= n_1 n_2 + n_2 n_3 + n_3 n_1 \end{align*}$
Given that:

$\begin{align*} m_1 + m_2 + m_3 &= n_1 + n_2 + n_3, \\ m_1 m_2 + m_2 m_3 + m_3 m_1 &= n_1 n_2 + n_2 n_3 + n_3 n_1, \\ m_1 m_2 m_3 &= n_1 n_2 n_3 \end{align*}$
are all true, Vieta's formulas imply that $P(x)$ and $Q(x)$ have the same roots. Setting combinations of roots equal to each other:

$\begin{align*} \sqrt{a + \frac{1}{b}} &= \sqrt{b + \frac{1}{a}} \implies a = b, \\ \sqrt{a + \frac{1}{b}} &= \sqrt{a + \frac{1}{c}} \implies b = c, \\ \sqrt{a + \frac{1}{b}} &= \sqrt{c + \frac{1}{b}} \implies a = c \end{align*}$
Thus, there will always be two equal sides in the triangle, proving that it is isosceles.

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#9 h Sep 5, 2024, 5:14 AM • 1 Y

Y by cubres

Here's my solution, the TeX is missing so I had to put the image

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cubres

#10 h Apr 12, 2025, 6:24 PM

Y by

Solution

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#11 h Apr 25, 2025, 3:44 PM

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WLOG, let $[ABC] = \frac12.$ Thus, $ah_A = bh_B = ch_C = 1.$ We can rewrite our equality as:
$\sqrt{a + \frac1b} + \sqrt{b + \frac1c} + \sqrt{c + \frac1a} = \sqrt{b + \frac1a} + \sqrt{c + \frac1b} + \sqrt{a + \frac1c}$ Let the 3 values on the left be $x, y, z,$ and the 3 values on the right be $X, Y, Z.$ Notice that:
$\begin{align*} x + y + z & = X + Y + Z \\ x^2 + y^2 + z^2 & = X^2 + Y^2 + Z^2 \implies xy + xz + yz = XY + XZ + YZ \\ xyz & = XYZ \end{align*}$ Thus, they are the roots to the same polynomial of degree 3. There exist 3 cases, if $x = X, Y, Z.$ In the first one, we can factor it to have that $(a - b)(ab + 1) = 0.$ Thus, $a = b.$ In the 2nd and 3rd, we can cancel the common term, and then $a = c$ or $b = c.$

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#12 h Yesterday at 4:34 PM

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We start by letting each term on the left-hand side $x_1,y_1,z_1$ respectively and each term on the right $x_2, y_2, z_2$ respectively. Then we have the obvious relations that
$\begin{align*} x_1+y_1+z_1 = x_2+y_2+z_2 \\ x_1^2+y_1^2+z_1^2 = x_2^2+y_2^2+z_2^2. \end{align*}$ We also claim that for $K=[ABC]$ ,
$\begin{align*} x_1y_1z_1 &= \sqrt{a+h_B}\sqrt{b+h_C}\sqrt{c+h_A} \\ &= \sqrt{abc+ab(h_C)+ac(h_B)+bc(h_A)+a(h_Bh_C)+b(h_Ah_C)+c(h_Ah_B)+h_Ah_Bh_C} \\ &= \sqrt{abc+2K\left(\dfrac{ab}{c}+\dfrac{ac}{b}+\dfrac{bc}{a}\right)+4K^2\left(\dfrac{a}{bc}+\dfrac{b}{ac}+\dfrac{c}{ab}\right)+h_Ah_Bh_C} \\ &= \sqrt{a+h_C}\sqrt{b+h_A}\sqrt{c+h_B} \\ &= x_2y_2z_2. \end{align*}$ In conjunction, these all imply that the polynomial with roots $x_1$ , $y_1$ , and $z_1$ is the same as the polynomial with roots $x_2$ , $y_2$ , and $z_2$ . This means that we have three cases.

Case 1: $\mathbf{x_1 = x_2}$ In this case, we have that $\sqrt{a+h_B} = \sqrt{a+h_C}$ , implying that $b = c$ .

Case 2: $\mathbf{x_1 = y_2}$ This time, we have that
$\begin{align*} \sqrt{a+h_B} &= \sqrt{b+h_A} \\ a+h_B &= b+h_A \\ a+\dfrac{2K}{b} &= b+\dfrac{2K}{a} \\ a^2b+2Ka &= ab^2+2Kb \\ (ab+2K)(a-b) &= 0 \end{align*}$ and since none of $a$ , $b$ , or $K$ can be zero, we must that $a=b$ .

\textbf{Case 3: $\mathbf{x_1 = z_2}$ } Finally, we have $\sqrt{a+h_B} = \sqrt{c+h_B}$ which explicitly gives that $a=c$ .

Thus, in all cases, $\triangle ABC$ is isosceles.

This post has been edited 1 time. Last edited by Kempu33334, Yesterday at 4:34 PM

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