Perpendicularity with incenter in a right triangle

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Perpendicularity with incenter in a right triangle

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Source: RMO Mumbai 2016, P1

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#1 h Oct 11, 2016, 7:14 AM • 1 Y

Y by Adventure10

Let $ABC$ be a right-angled triangle with $\angle B=90^{\circ}$ . Let $I$ be the incenter of $ABC$ . Draw a line perpendicular to $AI$ at $I$ . Let it intersect the line $CB$ at $D$ . Prove that $CI$ is perpendicular to $AD$ and prove that $ID=\sqrt{b(b-a)}$ where $BC=a$ and $CA=b$ .

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#2 h Oct 11, 2016, 7:22 AM • 1 Y

Y by Adventure10

Mixtilinear incircles!!!

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GoJensenOrGoHome

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GoJensenOrGoHome

#3 h Oct 11, 2016, 10:20 AM • 2 Y

Y by Adventure10, Mango247

WizardMath wrote:

Mixtilinear incircles!!!

You don't need such advanced things, since $\angle AID= \angle ABD=90$ we have $AIBD$ cyclic, so $\angle IAB= \angle IDB= \frac{\alpha}{2}$ and $\angle ABI= \angle ADI=45$ now we let $T=CI\cap AD$ and we have $\angle DTC= 180-\frac{\alpha+\gamma}{2}-45=180-45-45=90$ so $CI \perp AD$ .
By sine law $AD=\frac{b\cdot \sin(\gamma)}{\cos(\frac{\gamma}{2})}=2b\cdot \sin(\frac{\gamma}{2})=2b\sqrt{\frac{1-\cos (\gamma)}{2}}=b\sqrt{2(1-a/b)}=\sqrt{2b(b-a)}$ since $\frac{AD}{ \sqrt{2}}=DI$ we have the result

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#4 h Oct 11, 2016, 12:01 PM • 2 Y

Y by Adventure10, Mango247

can anyone draw the diagram.please.

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#5 h Oct 13, 2016, 3:35 AM • 3 Y

Y by Randomizeriitian, Adventure10, Mango247

GoJensenOrGoHome wrote:

WizardMath wrote:

Mixtilinear incircles!!!

You don't need such advanced things, since $\angle AID= \angle ABD=90$ we have $AIBD$ cyclic, so $\angle IAB= \angle IDB= \frac{\alpha}{2}$ and $\angle ABI= \angle ADI=45$ now we let $T=CI\cap AD$ and we have $\angle DTC= 180-\frac{\alpha+\gamma}{2}-45=180-45-45=90$ so $CI \perp AD$ .
By sine law $AD=\frac{b\cdot \sin(\gamma)}{\cos(\frac{\gamma}{2})}=2b\cdot \sin(\frac{\gamma}{2})=2b\sqrt{\frac{1-\cos (\gamma)}{2}}=b\sqrt{2(1-a/b)}=\sqrt{2b(b-a)}$ since $\frac{AD}{ \sqrt{2}}=DI$ we have the result

Nice solution. There is a small typo in your first line. It should be $\angle IAB = \angle IDC$ .

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MathematicalPhysicist

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MathematicalPhysicist

#6 h May 23, 2017, 4:32 PM • 1 Y

Y by Adventure10

GoJensenOrGoHome wrote:

WizardMath wrote:

Mixtilinear incircles!!!

You don't need such advanced things, since $\angle AID= \angle ABD=90$ we have $AIBD$ cyclic, so $\angle IAB= \angle IDB= \frac{\alpha}{2}$ and $\angle ABI= \angle ADI=45$ now we let $T=CI\cap AD$ and we have $\angle DTC= 180-\frac{\alpha+\gamma}{2}-45=180-45-45=90$ so $CI \perp AD$ .
By sine law $AD=\frac{b\cdot \sin(\gamma)}{\cos(\frac{\gamma}{2})}=2b\cdot \sin(\frac{\gamma}{2})=2b\sqrt{\frac{1-\cos (\gamma)}{2}}=b\sqrt{2(1-a/b)}=\sqrt{2b(b-a)}$ since $\frac{AD}{ \sqrt{2}}=DI$ we have the result

you actually dont need the law of sines.a simple manipulation of pythagoras' would do. anyways,a nice solution:)

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#7 h Mar 31, 2020, 9:04 AM • 3 Y

Y by amar_04, lilavati_2005, Purple_Planet

A complete synthetic non-trig solution, just for completeness

For the first part, observe that $\angle DBA=\angle DIA(=90^{\circ})$ , so $DBIA$ is cyclic, and hence, $\angle IAB=\angle IDB$ and $\angle ADI=\angle BI$ . So, $\angle ADB=45+\frac{A}{2}$ . Now in $\triangle DEC$ , $\angle DEC+\angle ECD+\angle CDE=180^{\circ}\implies \angle DEC=180^{\circ}-45^{\circ}-\frac{A}{2}-\frac{C}{2}=90^{\circ}$ , as desired.

For the second part, $AI=ID$ since $\angle ADI=45^{\circ}$ . Also, $CD=AC$ since $\triangle CED$ and $CEA$ are congruent by ASA criterion. So, $BD=CD-CB=CD-a=b-a$ .
Observe that $\triangle DBI$ and $\triangle AIC$ are similar, as $\angle BDI=\angle AIC=\frac{A}{2}$ and $\angle DIB=\angle CIA=\frac{C}{2}$ . So, $\frac{AI}{BD}=\frac{AC}{ID}\implies\frac{ID}{b-a}=\frac{b}{ID}\implies ID=\sqrt{b(b-a)}$ , as desired $\blacksquare$ .

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#8 h Jun 16, 2021, 11:01 AM

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Just use cosine rule and isosceles triangles for the 2nd part

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BVKRB-

#9 h Jun 26, 2021, 6:08 AM

Y by

Probably the most simple solution to this question with 0 trigonometry

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Diagram from Math-wiz :whistling:

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#10 h Nov 26, 2024, 11:22 AM

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Claim 1. $(ADBI)$ is cyclic.
$\angle AID=\angle AID=\angle ABD$ Claim 2. $AI=ID$
$\begin{align*}&45^{\circ}=\angle ABI=\angle ADI\\ &\angle DAI=180^{\circ}-\angle DBA-\angle IBA=45^{\circ}\end{align*}$ Claim 3. $\triangle CAD$ is isosceles $\implies CI\perp AD$
$\angle CAD=\angle CAI+\angle IAD=\angle IAB+45^{\circ}=\angle IDB+\angle ADI=\angle ADC$ Now we take $c=AB$ and we have the following
$ID=\sqrt{\frac{AD^2}{2}}=\sqrt{\frac{AB^2-DB^2}{2}}=\sqrt{\frac{c^2-(b-a)^2}{2}}=\sqrt{\frac{2b^2-2ba}{2}}=\sqrt{b(b-a)}$

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