Let ABC be a right triangle in B, with height BT, T on the hypotenuse AC. Construct the equilateral triangles BTX and BTY so that X is in the same half-plane as A with respect to BT and Y is in the same half-plane as C with respect to BT. Point P is the intersection of AY and CX. Show that PA⋅BC=PB⋅CA=PC⋅AB.
Problem
Source: Brazilian Mathematical Olympiad 2023, Level 3, Problem 2
Tags: geometry, right triangle, Equilateral Triangle
21.10.2023 23:00
Claim: Both TPYC and TPXA are cyclic. Proof: In fact, △TXC ~ △TAY since ∠XTC=∠ATY=150∘ and AT⋅TC=BT2=TX⋅TY⇒AT:TY=TX:TC. Hence, ∠PYT=∠PCT and ∠PXT=∠PAT, as we wanted. Now, we have CP⋅CX=CT⋅CA=CB2⇒CB is tangent to (BPX)⇒∠CBP=∠BXC. By the Cevian Ratio Lemma in △CBX, PXPC=BXBC⋅sin∠XBPsin∠BXP=BX⋅PXBC⋅BP⇒BPPC=BXBC=BTBC=ABAC. The other ratios are completely analogous and we're done.
21.10.2023 23:07
Construct A1 outwards from ABC such that BCA1 is equilateral, and notice that quadrilaterals ATYB and ABA1C are similar (because ABT∼ACB and BTY∼CBA1). Another way to think of this is that they can be transformed into each other by a homothety with center A and a reflection through the bisector of ∠BAC. This similarity gives us ∠CAY=∠BAA1. Doing an analogous process on cevian CX, we notice that P is the isogonal conjugate of the Fermat Point of ABC (The point F for which ∠AFB=∠AFC=∠BFC=120∘). The isogonal conjugate of the Fermat Point is the first isodynamic point of ABC which indeed satisfies the equations. But, if you didn't know this, you could also extend PA, PB and PC until they intersected (ABC) at A′, B′ and C′. Then: ∠A′B′C′=∠A′B′B+∠BB′C=∠BAA′+∠BCC′=∠FAC+∠FBC=60∘ We repeat the same process on the other two angles of A′B′C′ to conclude it's equilateral. Lastly, from BPC∼C′PB′ and APB∼B′PA′, we get: BCCP=B′C′PB′=A′B′PB′=ABAP⟹BC⋅AP=AB⋅CP Repeating this for another pair of sides yields the conclusion.
23.10.2023 19:00
voce pode mandar o link da prova por favor
26.10.2023 06:21
Analytic geometry bash: Let T=(0;0),A=(a;0),B=(0;b),C=(c;0). WLOG a<0,b>0,c>0. Now, ΔABT ~ ΔBCT⇔|AT||BT|=|BT||CT|⇒b2=−ac. Now, X=(x;b2) and |XT|=|BT|⇒√x2+b24=√b2⇒ x2=3b24. Given that b>0 and X is on the same halfplane of A wrt the y axis, X=(−b√32;b2). Similarly, Y=(b√32;b2). The equation of the line through A and Y is y=bb√3−2ax−abb√3−2a. Similarly, the equation of the line through C and X is y=−bb√3+2cx+bcb√3+2c. Therefore, P=((a+c)b√32b√3+2c−2a;b(c−a)2b√3+2c−2a). Now, it is enough to prove that |PA|⋅|BC|=|PB|⋅|AC|, then the other equality is similar. |PA|⋅|BC|=|PB|⋅|AC|⇔ √((a+c)b√32b√3+2c−2a−a)2+(b(c−a)2b√3+2c−2a)2⋅√b2+c2= √((a+c)b√32b√3+2c−2a)2+(b(c−a)2b√3+2c−2a−b)2⋅√(c−a)2⇔ (((c−a)(b√3−2a))2+(b(c−a))2)(b2+c2)= (((a+c)b√3)2+(b(a−c−2b√3))2)(c−a)2⇔ (3b2−4ab√3+4a2+b2)(b2+c2)= b2(3a2+6ac+3c2+a2+c2+12b2−2ac−4ab√3+4bc√3)⇔ (a2+b2−ab√3)(b2+c2)=b2(a2+c2+ac+3b2−ab√3+bc√3)⇔ a2b2+a2c2+b4+b2c2−ab3√3−abc2√3= a2b2+b2c2+ab2c+3b4−ab3√3+b3c√3⇔ a2c2−abc2√3=ab2c+2b4+b3c√3. But b2=−ac⇒a2c2=b4,ab2c=−b4,−abc2√3=b3c√3, as desired.
26.10.2023 06:24
Genius132 wrote: voce pode mandar o link da prova por favor https://www.obm.org.br/content/uploads/2023/10/Nivel_3_OBM_2023.pdf
31.10.2023 22:07
Let T=(0;0),A=(a;0),B=(0;2),C=(−c;0),Y=(√3,1),X=(−√3,1). Then, ac=4. Let P(x;y). Then P∈AY⇒yx+a=1√3+a and P∈CX⇒yx−c=−1√3+c, yielding: (x+a)(√3+c)=(c−x)(√3+a)x=√3(c−a)2√3+a+cWriting d for this denominator, we get: x+a=√3(c+a)+a(a+c)d=(a+c)(√3+a)dSo c−x=(a+c)(√3+c)d, y=a+cd, and 2−y=a+c+4√3d. Now, using the distance formula, we get: PA2⋅BC2⋅d2=d2⋅((x+a)2+y2)⋅(c2+4)Since c2+4=c2+ac=c(c+a) and we have a (c+a)2 common term in both (x+a)2 and y2, we can write: PA2⋅BC2⋅d2=(a+c)3⋅((a+√3)2+1)⋅c=(a+c)3⋅(4a+4c+8√3)As this expression is symmetric in a and c, it will also be the same for PC2⋅BA2⋅d2. Finally, we have: PB2⋅AC2⋅d2=d2⋅(x2+(2−y)2)⋅(c+a)2=(a+c)2⋅(3(c−a)2+(a+c+4√3)2)=(a+c)2⋅(4a2+4c2−4ac+8√3(a+c)+48) Writing 48=12ac, enables the final factoring PB2⋅AC2⋅d2=d2⋅(4(a+c)2+8√3(a+c))=(a+c)3⋅(4a+4c+8√3)
24.08.2024 18:04
Let Y′ in the exterior of side BC such that ΔBY′C is equilateral. Because ΔABC is rightangle, we get ΔBAT∼ΔCAB⇒BATY∼CABY′⇒∠BAY′=∠CAY. Defining X′ analogously and F=AY′∩CX′ we conclude that P is the isogonal conjugate of F, but F is the fermat point of ΔABC⇒P is an isodynamic point, and that gives us the desired equality.
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