Find the smallest positive real number

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Find the smallest positive real number

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Source: IMO Shortlist 2007, G6 / USA TST 2008, Day 2, Problem 6

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#1 h Jul 13, 2008, 2:06 AM • 4 Y

Y by Davi-8191, Adventure10, Mango247, and 1 other user

Determine the smallest positive real number $k$ with the following property. Let $ABCD$ be a convex quadrilateral, and let points $A_1$ , $B_1$ , $C_1$ , and $D_1$ lie on sides $AB$ , $BC$ , $CD$ , and $DA$ , respectively. Consider the areas of triangles $AA_1D_1$ , $BB_1A_1$ , $CC_1B_1$ and $DD_1C_1$ ; let $S$ be the sum of the two smallest ones, and let $S_1$ be the area of quadrilateral $A_1B_1C_1D_1$ . Then we always have $kS_1\ge S$ .

Author: Zuming Feng and Oleg Golberg, USA

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Vrajitoarea Andrei

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Vrajitoarea Andrei

#2 h Oct 12, 2008, 9:19 PM • 1 Y

Y by Adventure10

This is from IMO Shorlist 2007,problem G6

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#3 h May 29, 2013, 2:00 AM • 4 Y

Y by Mediocrity, 277546, Adventure10, Mango247

The answer is $k=1$ . We can get $\frac{S}{S_1}$ arbitrarily close to $1$ by taking $AA_1BB_1CC_1DD_1$ arbitrarily close to a medial triangle configuration (e.g. approaching $D_1=A=A_1$ ).

Lemma. Let $\triangle{XYZ}$ be a triangle and $X_1,Y_1,Z_1$ points on sides $YZ$ , $ZX$ , $XY$ . Then $[X_1Y_1Z_1] \ge \min_{\text{cyc}}([XY_1Z_1])$ .

Proof. For convenience, take an affine transformation sending $\triangle{X_1Y_1Z_1}$ to an equilateral triangle, and construct its antimedial triangle $\triangle{X_2Y_2Z_2}$ . Let $O$ be the circumcenter of $\triangle{X_1Y_1Z_1}$ .

Suppose the ``border areas'' $[XY_1Z_1],\ldots$ are all strictly greater than $[X_1Y_1Z_1]$ . Then $d(X,Y_1Z_1) > d(X_1,Y_1Z_1)$ , so $X$ cannot lie inside $(OY_1X_2Z_1)$ . Since $Y_1Z_1$ separates $X,X_1$ , we must have $\angle{Y_1XZ_1}<60^\circ$ . By symmetry, we obtain $\angle{X}+\angle{Y}+\angle{Z} < 3\cdot60^\circ=180^\circ$ , contradiction. $\blacksquare$

We will now prove $S_1 \ge S$ . Inspired by the degenerate case, we define $T = A_1C_1\cap B_1D_1$ , $f(A) = [AA_1D_1]$ , $g(A) = [TA_1D_1]$ , and $h(A) = \max([B_1A_1D_1],[C_1A_1D_1])$ ; similarly define $f(X),g(X)$ for $X=B,C,D$ .

Case 1: There exists $X\in\{A,B,C,D\}$ such that $f(X) \le h(X)$ . WLOG $f(A) \le h(A)$ and $[B_1A_1D_1] \le [C_1A_1D_1]$ . Then
$\begin{align*} S_1-S &\ge [C_1A_1D_1]+[A_1B_1C_1]-f(A)-\min(f(B),f(C),f(D)) \\ &\ge [A_1B_1C_1]-\min(f(B),f(C),f(D)). \end{align*}$ But $A_1BCDD_1$ is convex, so letting $B' = CB\cap D_1A_1$ and $D' = CD\cap A_1D_1$ , we find $f(B) \le [B'A_1B_1]$ , $f(C) = [CB_1C_1]$ , $f(D) \le [D'C_1D_1] \le [D'C_1A_1]$ . The lemma now immediately gives $S_1-S \ge0$ , as desired.

Case 2: $f(X)>h(X)>g(X)$ for every $X\in\{A,B,C,D\}$ , so $d(A,A_1D_1)>d(T,A_1D_1)$ , etc. Taking an affine transformation if necessary, WLOG assume $\angle{A_1TB_1}=90^\circ$ . For convenience, we write $T=(0,0)$ , $A_1=(-a,0)$ , $B_1=(0,b)$ , $C_1=(c,0)$ , $D_1=(0,-d)$ for positive reals $a,b,c,d$ .

Then in particular, either $x(A)<-a$ or $y(A)<-d$ must hold (analogous statements hold for $B,C,D$ ). WLOG suppose $y(A)<-d$ . Then we easily deduce
$\begin{align*} y(D)>-d &\implies x(D)>c>x(C) \\ &\implies y(C)>b>y(B)\implies x(B)<-a<x(A). \end{align*}$ A simple ratio calculation (note $A$ is bounded by $C_1D_1$ , $D_1A_1$ , and $x=-a$ ) yields
$[C_1A_1D_1] \le h(A) < f(A) \le \frac{a}{c}[A_1D_1C_1],$ whence $a>c$ . But similarly, we deduce $b>d$ , $c>a$ , and $d>b$ , contradiction.

Comment. There are certainly other ways to prove the lemma. Most obviously, we can directly use barycentric coordinates or standard area calculations. If we want to avoid both affine transformations and calculation, we can compare $\triangle{X_1Y_1Z_1}$ with the medial triangle $\triangle{X'Y'Z'}$ of $\triangle{XYZ}$ .

Comment. The key is to look at degenerate cases (and realize the answer is $k=1$ ). "Simple case analysis" is one way to motivate this, but smoothing provides a bit more insight, both for why the degenerate case should be optimal and how to go about proving $S_1\ge S$ . Indeed, suppose we have a configuration with $S/S_1$ maximal (or at least ``close'' to maximal). Then we can smooth to the case $f(A)\le f(B)=f(D)\le f(C)$ (otherwise, consider rotating $BC,CD$ about $B_1,C_1$ to increase $S$ while fixing $S_1$ ). Now if $f(A)<f(B)=f(D)$ and $B_1$ is closer to $AB$ than $D_1$ (or $C_1$ is closer to $AD$ than $A_1$ ), we can slide $A_1$ away from $A$ a bit (resp. $D_1$ away from $A$ ) to decrease $S_1$ while increasing $S$ .

Hence an ``optimal'' configuration (with $S/S_1$ maximal) should have $d(D_1,AB) \le d(B_1,AB)$ (and $d(A_1,AD) \le d(C_1,AD)$ ). While we can't do much more obvious smoothing, this at least suggests we consider the case when at least one of $A_1,D_1$ coincides with $A$ (or just $f(A)$ is really small).

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#4 h Apr 17, 2015, 5:30 AM • 4 Y

Y by stroller, Modesti, Adventure10, Mango247

Lemma: in a triangle ABC with X,Y,Z on sides BC,CA,AB then (XYZ) $\ge$ (AYZ),(BXZ) or (CXY)
Proof: WLOG ABC is quadrilateral, now just computations.

The proof k=1 is easy with this result.WLOG rays AB and DC intersect at P, then apply lemma to triangle APD and we get S1 > (A1C1D1) $\ge$ (AA1D1), (DDAC1) or (PA1C1). In the last case, (PA1C1) > S so we are done. Otherwise, WLOG (A1C1D1) $\ge$ (DC1D1) and we easily get (A1B1C1) > (A1BB1) or (B1CC1), so we get S1 > S.

For equality let D=A=D1, and A1, B1, C1 be midpoints of ABC.

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awesomeming327.

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awesomeming327.

#5 h Oct 12, 2023, 12:55 AM

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We claim that the answer is $1$ . First, we'll show that $S_1=S$ can be achieved. Let $ABC$ be a triangle, and $D$ be infinitely close to the midpoint of $AC$ . Then, we can let $A_1B_1C_1D_1$ essentially be equivalent to the medial triangle of $\triangle ABCD$ , giving an answer of $1$ .

Now, we show that $S_1\ge S$ is true. We claim that for any triangle $XYZ$ and $X_1$ on $YZ$ , $Y_1$ on $ZX$ , and $Z_1$ on $XY$ , $[X_1Y_1Z_1]\ge \text{min}([XY_1Z_1],[YZ_1X_1],[ZX_1Y_1])$ Assume otherwise, then apply the affine transform that will take $[X_1Y_1Z_1]$ to an equilateral triangle. Let $X'$ , $Y'$ and $Z'$ be such that $\triangle X_1Y_1Z_1$ is the medial triangle of $\triangle X'Y'Z'$ . Evidently, $X'$ is closer to $Y_1Z_1$ than $X$ . Since $X'Y_1Z_1$ is equilateral, that means $X$ lies outside $(X'Y_1Z_1)$ so $\angle YXZ=\angle Y_1XZ_1>60^\circ$ . Similarly, $\angle XZY,\angle ZYX>60^\circ$ , contradiction because the sum of angles in a triangle adds to $180^\circ$ . This proves our claim.

Now, let $E$ be $AB$ intersection with $CD$ . Without loss of generality, $BC$ is closer to $E$ than $AD$ . Let $T_1=[AA_1D_1]$ , $T_2=[BB_1A_1]$ , $T_3=[CC_1B_1]$ , $T_4=[DD_1C_1]$ then by our claim,
$\begin{align*} S_1 &= [A_1C_1D_1] + [A_1B_1C_1] \\ &= \text{min}(T_1,T_2+T_3,T_4)+\text{min}(T_3,T_1+T_4,T_2) \\ &\ge S \end{align*}$ as desired.

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#6 h Feb 23, 2024, 4:31 AM

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The answer is $k=1$ . For a construction, consider the degenerate case where $ABCD$ is a triangle $ABC$ with $D$ the midpoint of $\overline{BC}$ , where $A_1B_1C_1D_1$ are the midpoints of their respective sides (hence forming the medial triangle of $ABC$ .)

To show that this is minimal, consider the following:

Claim: For any triangle $DEF$ inscribed on the sides of a triangle $ABC$ , $[DEF] > \operatorname{min}([AEF], [BDF], [CDE])$ .

Proof. Take an affine transform sending $DEF$ to an equilateral triangle. Assume for the sake of contradiction that the result is not true; then, let $XYZ$ be the image of $DEF$ upon a homothety of ratio $4$ at its center. It is equivalent to show that there cannot exist points $A, B, C$ outside $XYZ$ such that $DEF$ is inscribed in $ABC$ .

However, notice that for any point $A$ outside $XYZ$ , we always have $\angle EAF < 60^\circ$ (as the circle for which $\angle EAF = 60^\circ$ is tangent to triangle $XYZ$ ). Hence the angles of $ABC$ add up to less than $180^\circ$ , contradiction. $\blacksquare$

Now WLOG $BC < AD$ and construct the point $E = \overline{AB} \cap \overline{CD}$ . The lemma implies that $[B_1A_1C_1] < \operatorname{min}([BA_1B_1], [CB_1C_1])$ and $[A_1D_1C_1] < \operatorname{min}([DC_1D_1], [AA_1D_1])$ , and summing yields the result.

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