Altitude drawn to the hypotenuse

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Altitude drawn to the hypotenuse

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Source: IMO 1988/5, IMO Shortlist 13, IMO Longlist 23

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orl

#1 h Oct 22, 2005, 3:20 PM • 2 Y

Y by Adventure10, Mango247

In a right-angled triangle $ABC$ let $AD$ be the altitude drawn to the hypotenuse and let the straight line joining the incentres of the triangles $ABD, ACD$ intersect the sides $AB, AC$ at the points $K,L$ respectively. If $E$ and $E_1$ dnote the areas of triangles $ABC$ and $AKL$ respectively, show that
$\frac {E}{E_1} \geq 2.$

This post has been edited 3 times. Last edited by orl, Sep 13, 2008, 1:05 AM

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#2 h May 13, 2006, 1:59 PM • 2 Y

Y by Adventure10, Mango247

Lemma: Through the incenter $I$ of $\triangle{ABC}$ draw a line that meets the sides $AB$ and $AC$ at $P$ and $Q$ , then:
$\frac{AB}{AP} \cdot AC + \frac{AC}{AQ} \cdot AB = AB+BC+AC$
Proof of the lemma:
Consider the general case: $M$ is any point on side $BC$ and $PQ$ is a line cutting AB, AM, AC at P, N, Q. Then:

$\frac{AM}{AN}=\frac{S_{APMQ}}{\triangle{APQ}}=\frac{\triangle{APM}+\triangle{AQM}}{\triangle{PQA}}=\frac{\frac{AP}{AB}\triangle{ABM}+\frac{AQ}{AC}\triangle{ACM}}{\frac{AP\cdot AQ}{AB \cdot AC}}=$

$=\frac{AC}{AQ}\cdot \frac{BM}{BC}+\frac{AB}{AP}\cdot \frac{CM}{BC}$

If $N$ is the incentre then $\frac{AM}{AN}=\frac{AB+BC+CA}{AB+AC}$ , $\frac{BM}{BC}=\frac{AB}{AB+AC}$ and $\frac{CM}{BC}=\frac{AC}{AC+AB}$ . Plug them in we get:
$\frac{AB}{AP} \cdot AC + \frac{AC}{AQ} \cdot AB = AB+BC+AC$

Back to the problem
Let $I_1$ and $I_2$ be the areas of $\triangle{ABD}$ and $\triangle{ACD}$ and $E$ be the intersection of $KL$ and $AD$ . Thus apply our formula in the two triangles we get:
$\frac{AD}{AE} \cdot AB + \frac{AB}{AK} \cdot AD = AB+BD+AD$
and
$\frac{AD}{AE} \cdot AC + \frac{AC}{AL} \cdot AD = AC+CD+AD$
Cancel out the term $\frac{AD}{AE}$ , we get:
$\frac{AB+BD+AD-\frac{AB}{AK} \cdot AD }{AC+CD+AD- \frac{AC}{AL} \cdot AD }=\frac{AB}{AC}$
$AB \cdot CD + AB \cdot AD - \frac{AB \cdot AC \cdot AD}{AL}=AC \cdot BD+ AC \cdot AD -\frac{AB \cdot AC \cdot AD}{AK}$
$AB+AB \cdot \frac{CD}{AD}-\frac{AB \cdot AC}{AL}=AC+ AC \cdot \frac{BD}{AD} - \frac{AB \cdot AC}{AK}$
$AB+AC - \frac{AB \cdot AC}{AL}=AB+AC - \frac{AB \cdot AC}{AK}$
$\frac{AB \cdot AC}{AK} = \frac{AB \cdot AC}{AL}$
So we conclude $AK=AL$ .

Hence $\angle{AKI_1}=45^o=\angle{ADI_1}$ and $\angle{ALI_2}=45^o=\angle{ADI_2}$ , thus $\triangle{AK_1} \cong \triangle{ADI_1}$ and $\triangle{ALI_2} \cong \triangle{ADI_2}$ . Thus $AK=AD=AL$ . So the area ratio is:
$\frac{E}{E_1}=\frac{AB \cdot AC}{AD^2} = \frac{BC}{AD} =\frac{BD+CD}{\sqrt{BD \cdot CD}}\geq 2$

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probability1.01

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probability1.01

#3 h Jul 26, 2006, 4:51 AM • 1 Y

Y by Adventure10

I think we can be a bit quicker than that. Let X and Y be respective incenters of $ABD$ and $ACD$ . Note that by spiral similarity taking $BDA$ to $ADC$ , we have $XDY$ being a 45 degree rotation and dilation of both $BDA$ and $ADC$ (of course rotating in opposite directions). It follows then that $AK = AL$ . Then $\angle AKX = 45 = \angle ADX \implies AK = AD$ . Similarly, $AL = AD$ . Then $\frac{[ABC]}{[AKL]}= \frac{AB \cdot AC}{AK \cdot AL}= \frac{AB \cdot AC}{AD^{2}}$ . It's an easy finish from here; as shobber did, we can note that $BC \ge 2AD$ , and so $\frac{AB \cdot AC}{AD^{2}}= \frac{BC}{AB}\cdot \frac{AB}{AD}= \frac{BC}{AD}\ge 2$ .

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#4 h Jul 26, 2006, 11:05 PM • 2 Y

Y by Adventure10, Mango247

Remark. Given are a triangle $ABC$ and a point $D\in [BC]$ . Denote the incircles $C(I_{1},r_{1})$ , $C(I_{2},r_{2})$ of the triangles $ABD$ , $ACD$ respectively and the points $M\in AB\cap DI_{1}$ , $N\in AC\cap DI_{2}$ , $X\in AB\cap I_{1}I_{2}$ , $Y\AC\cap I_{1}I_{2}$ . Prove that $XY\parallel MN\Longleftrightarrow AX=AY$ .

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orl

#5 h Aug 12, 2008, 10:02 PM • 1 Y

Y by Adventure10

Remark by 28121941:

This was problem 5 at IMO 1988, proposed by Greece. As (ABC) (area) is greater or equal 2 times (AKL), the minimal value of the quotient is 2.
At Gazeta Matematica 1991(number 10), Neculai Roman published two generalizations of this problem:
1) Let ABC a triangle. Let k_1 the circle through A and B and tangent to AC; analogously, let k_2 the circle through A and C and tangent to AB. The second intersection of k_1 and k_2 is D. The line defined by the incenters of the triangles ABD and ACD meet the lines AB,AC in K and L, respectively. If S is the area of ABDC and T is the area of ACD, show that S is greater or equal to 2 times T.

2) Let ABC a triangle. with A>B, A>C. let D and D' two points of the segment BC such that angle CAD = angle ABC and angle BAD' = angle ACB. The line which join the incenters of ABD and ACD' intersect AB at K and AC at L.
If S = area(ABC) and T = area(AKL), then S is greater or equal to 4T*(sin(A/2))^2.

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#6 h Dec 18, 2019, 7:47 PM • 5 Y

Y by AlastorMoody, GeoMetrix, amar_04, Adventure10, Mango247

Here's a thoughtless bary bash

orl wrote:

In a right-angled triangle $ABC$ let $AD$ be the altitude drawn to the hypotenuse and let the straight line joining the incentres of the triangles $ABD, ACD$ intersect the sides $AB, AC$ at the points $K,L$ respectively. If $E$ and $E_1$ dnote the areas of triangles $ABC$ and $AKL$ respectively, show that
$\frac {E}{E_1} \geq 2.$

Set $ABC$ as reference triangle.Let $P=AI_1\cap BC$ and $Q=AI_2\cap BC$ .Then clearly we have $P=(0:b:a-b)\implies I_1=(a(a-b):bc:c(a-b))$ $Q=(0:a-c:c)\implies I_2=(a(a-c):b(a-c):bc)$ Now let $K=(x:y:0)$ .Then we have $\left|\begin{array}{ccc}x & y& 0 \\ a(a-b) & bc & c(a-b) \\ a(a-c) & b(a-c) & bc\end{array}\right|=0\implies K=(a-b:b:0)$ Similarly $L=(a-c:0:c)$ .Therefore $\dfrac{[AKL]}{[ABC]}=\left|\begin{array}{ccc}1 & 0 & 0 \\ 1-\tfrac{b}{a} & \tfrac{b}{a} & 0 \\ 1-\tfrac{c}{a} & 0 & \tfrac{c}{a}\end{array}\right| = \dfrac{bc}{a^2}=\dfrac{bc}{b^2+c^2}\leq \dfrac{1}{2} \; \square$ Thus we are done.

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