For triangles with equal inradii

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For triangles with equal inradii

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#1 h Jan 18, 2017, 2:51 AM • 3 Y

Y by PRO2000, jonlin1000, Adventure10

Four points lie on a plane such that no three of them are collinear. Consider the four triangles formed by taking any three points at a time. If the inradii of these four triangles are all equal, prove that the four triangles are congruent.

This post has been edited 1 time. Last edited by Ankoganit, Jan 18, 2017, 3:08 AM

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#2 h Apr 7, 2017, 11:29 AM • 2 Y

Y by Adventure10, Mango247

The solution I submitted:
$\textbf{Solution}$ :
Let the quadrilateral be $ABCD$ with the points lying in this order and that any side doesn't cross any other.
Assume that the quadrilateral $ABCD$ is concave, without loss of generality let it be concave at $D$ . Then since the incircle is the largest circle that is completely inside a triangle, and since $D$ lies strictly inside $\triangle ABC$ , the inradius of $ABC$ is strictly greater than that of $BDC$ , which is contrary to our assumption. Thus the quadrilateral is convex.
By $[F]$ we denote the area of a figure $F$ , by $s_{MNP}$ we denote the semiperimeter of the triangle $MNP$ , by $r_{MNP}$ we denote the inradius of the triangle $MNP$ .
Note that by the condition, $[ABC]/s_{ABC}=[ABD]/s_{ABD}=[DBC]/s_{DBC}=[ADC]/s_{ADC}=[ABCD]/(s_{ABC}+s_{ACD})=[ABCD]/(s_{ABD}+s_{BCD})$ , which implies that the sum of the semiperimeters of triangles $ABC$ and $ACD$ is the same as that of $ABD$ and $BCD$ .
$\Longrightarrow AB + BC + CA + AC + CD + DA = AB + BD + DA + BC + CD + DB \Longrightarrow AC = BD$ .
Now we show that $\triangle ADC \cong \triangle BCD$ , from which the others follow similarly.
Reflect $A$ about the perpendicular bisector of $CD$ to $B'$ . Draw the semicircle $(D)$ with center as $D$ and radius $DB'$ , with $CD$ lying on one of the diameters. Clearly $B$ should be on $(D)$ because $DB= AC= DB'$ . Suppose $B \neq B'$ .
Select an arbitrary point $X$ on the $(D)$ . The inradius of $XDC$ is given by
$r_{XDC}=\frac{m \sin x}{k+\sqrt{k^2-2m(1+ \cos x)}}$ where $m = XD.DC = BD.DC$ , $k = XD + DC = BD + DC$ , both of which are constant, and $x=\angle XDC$ .
Let $AC$ intersect $(D)$ at $J$ . We claim that as a function of $x$ there is atmost one X such that $f'(x)=0$ where the function is $r_{XDC}$ as $X$ varies from ray $DC \cap (D) = Y$ to $J$ , excluding $Y$ , (noting that the function is continuous and differentiable in the interval), for then, either the inradius increases from $B'$ to the maximum and then decreases till $J$ or the inradius increases till $J$ . In either case, the inradius of $B'DC$ is not the inradius of any $\triangle XBC$ , because the inradius strictly decreases in the second case, and in the first case because the radius can't decrease as much to equal the inradius of $B'DC$ because the inradius of $JDC$ is more than that of $B'DC$ as $B'DC$ is inside $JDC$ , and due to the condition we prove, because otherwise it would increase after decreasing, which would violate the condition, for the slope of the tangent of the graph of the function will be 0 again after becoming 0 once. This analysis would imply that $B$ would not lie in the interval $(Y,J)$ , and thus that $AC$ and $BD$ intersect outside the quadrilateral, which would be a contradiction, because it would violate the ordering we gave to the quadrilateral, and thus leading to a contradiction to our assumption that $B \neq B'$
Now we prove that there is exactly one $x$ in the given interval such that $f'(x)=0$ . $f'(x)=0$ is equivalent to
$mk \cos x +m \cos x \sqrt{k^2-2m(1+ \cos x)} = \frac {m^2 \sin^2 x}{\sqrt{k^2-2m(1+ \cos x)}}$ Note that $x$ can't be in the interval $[\pi/2,\pi]$ because the LHS would otherwise be negative or 0, while the RHS is not at the same time. So the only such possibility is when $x \in (0,\pi/2)$ .
On substituting $\cos x=t$ , the equation becomes $mt^2+t(2m-k^2)-tk\sqrt{k^2-2m(1+t)}+m=0$ . Define $P(t)=mt^2+t(2m-k^2)-tk\sqrt{k^2-2m(1+t)}+m$ . This is clearly continuous. When $x=0$ , $t=1$ , and $P(1)=-(BD-DC)^2-(BD+DC)|BD-DC| \leq 0$ but when $BD=DC$ , $x\ne 0$ because then the triangle degenerates. When $x=\pi/2$ , $t=0$ , and $P(0) = m>0.$ So there exists an odd number of solutions for acute $x$ , because in the graph of $f'(t)$ , the places where it cuts the $x$ axis are precisely the points of change of sign. Since the equation on separating the expression with the radical on the other side and squaring yields a quartic expression which can have at most 4 real roots, there can be either 1 or 3 roots for which $x$ is acute. In particular, the expression is $m^2t^4 + 4m^2 t^3 + \_ t^2 + \_ t + m^2=0$ where the $\_$ stand for some coefficients we don't have to work with. From here, we can see that 3 roots don't exist for which $x$ is acute, because when we expand the quartic, we observe that the sum of roots is -4, and the product of the roots is 1, from which if there were 3 real roots, then there would have been 4 real roots, as complex roots occur in conjugates, and at least 2 roots must be negative since product is 1 and sum is negative, which gives us $\leq$ 2 roots for which $x$ is in the interval $(0,\pi/2)$ , which is a contradiction, which leads to $B=B'$ and thus to the fact that $\triangle ABD \cong \triangle ADC$ due to the reflection. Similarly we show that the other pairs of triangles having the same base (as one of the sides of the quadrilateral) are congruent, and due to the transitivity of congruences, we conclude that all triangles are congruent. \ \ \ $\blacksquare$
$\textbf{Note}:$ We can also prove the fact that $ABCD$ is a rectangle using the conclusion of the solution.

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#3 h Apr 7, 2017, 11:44 AM • 2 Y

Y by Adventure10, Mango247

This is also Japan MO 2011 P5.

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#4 h Jan 13, 2020, 5:56 PM • 1 Y

Y by Adventure10

An alternative approach:
We know that [ABC]+[ACD]=[BAD]+[BCD]....using [area]= $s*r$ here, we get that the diagonals are equal,lets call them x.
Using s*r=( (s)(s-a)(s-b)(s-x) )^1\2 for any one of the triangles where s=(a+b+x)/2 and a and b are two corresponding adjacent sides and simplifying gives us a cubic in x: $x^3-x^2(b+c)-x(a^2+b^2-2ab+(a+b)r^2)+(a+b)( (a-b)^2-r^2 )$
We know that x is same for any two sides=> cubic has same coefficients wrt any 2 adjacent sides. So,the cubic reveals that sum of adjacent sides is equal. Which implies that opposite sides of ABCD are equal. :jump:

:jump:

As two adjacent sides and third side(diagonal) are equal, the triangles are congruent by SSS test.

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