Given is a triangle ABC with altitude AH, diameter of the circumcircle AD and incenter I. Prove that ∠BIH=∠DIC.
Problem
Source: St Petersburg 2022 10.3
Tags: geometry
29.09.2022 00:17
Excellent problem for practicing application of the following lemma: If 0∘<x,y<α<180∘ such sinxsin(α−x)=sinysin(α−y) then x=y. Note that sin∠BIHsin∠BHI=BHBI, sin∠CIHsin∠CHI=CHCI so sin∠BIHsin∠CIH=BH⋅CICH⋅BI. We know that BHAH=cotβ, CHAH=cotγ, BICI=sinβ2sinγ2. On the other side, we know that ∠IBD=90∘−β2, ∠ICD=90∘−γ2. Now we have sin∠CIDsin∠ICD=CDID, sin∠BIDsin∠IBD=BDID. This implies sin∠CIDsin∠BID=CD⋅cosβ2BD⋅cosγ2. Also, we know that CDBD=cosβcosγ. The rest is now a simple calculation.
29.09.2022 22:22
Consider inverting in A swapping B,C. Is known I goes to the A-excenter J,D goes to H, and said M the midpoint of arc BC, we have: ∠AHI=∠AJD=∠JID, the first is just inversion, the second is becouse ∠AMD=π/2, and is known M is the midpoint of IJ, so now ∠BIH=−β/2+90−∠AHI,∠DIC=∠MIC−∠JID=90−β/2−∠AHIQ. E. D.
25.10.2022 15:14
Angle divider theorem From #2, we only need to show sin∠BIHsin∠HIC=sin∠CIDsin∠BID. From the angle devider theorem, left side =BHHC×CIBI=ABcosBACcosC×sinB2sinC2=sinCcosBsinBcosC×sinB2sinC2=cosBcosC2cosCcosB2.Applying Ceva theorem to point P and △BIC, sin∠IBDsin∠CBD×sin∠BCDsin∠ICD×sin∠CIDsin∠BID=1.Notice that ∠CBD=∠CAD=π2−∠ADC=π2−B, so∠IBD=∠IBC+∠CBD=B2+π2−B=π2−B2,Similarly, ∠ICD=π2−C2,∠BCD=π2−C. Applying the induction formula, plugging into (2), cosB2cosB×cosCcosC2×sin∠CIDsin∠BID=1.Lastly, sin∠BIHsin∠HIC=sin∠CIDsin∠BID=cosBcosC2cosCcosB2.
27.10.2022 13:00
I proposed this problem and here you could find three solutions (in Russian): https://youtu.be/6k4Sgb8fmJQ First one is angle chasing, second is trigonimetric and third one is with harmonic quadruples and isogonal conjugation with respect to the quad. Also, as it was mentioned by Michael Greenberg this problem was posted previously on aops: https://artofproblemsolving.com/community/c6h2249136p17322840
20.12.2022 15:44
I have seen this problem 7 years ago...