Math Olympiad | A Very Nice Geometry Problem | 2 Different Methods
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- Опубліковано 3 жов 2024
- Math Olympiad | A Very Nice Geometry Problem | 2 Different Methods
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I did it a different way (also without trigonometry):
Extend line AC to a point D, such that line BD ⊥ AD and ∠ADB = 90°
As ∠ADB = 90° and ∠DAB = 45° ⇒ ∠ABD is also 45° (angles of a triangle)
As ∠DAB = ∠ABD = 45° ∴ △ABD is an isosceles right-triangle (45°-45°-90°) with BD = AD
Now, ∠ABC = 22.5° (given) and as ∠ABD = 45° ⇒ ∠CBD = ∠ABD - ∠ABC = 45° - 22.5° = 22.5°
∴ ∠CBD = ∠ABC = 22.5°, making BC the angle bisector of ∠ABD for △ABD
If we let AC = 𝑥 and CD = 𝑦, as AD = BD (isosceles triangle) ⇒ BD = 𝑥 + 𝑦
⇒ AB = √2(𝑥 + 𝑦), as AB² = BD² + AD² = 2(𝑥 + 𝑦)² (Pythagoras' Theorem)
⇒ AB/AC = BD/CD (angle-bisector theorem)
⇒ (𝑥 + 𝑦)/𝑦 = √2(𝑥 + 𝑦)/x
⇒ 1/y = √2/𝑥
∴ 𝑦 = 𝑥/√2 (CD = AC/√2)
Now, taking △BCD, as ∠CDB = 90° ⇒ CD² + BD² = BC²
⇒ 𝑦² + (𝑥 + 𝑦)² = 16
As 𝑦 = 𝑥/√2 ⇒ 𝑥²/2 + (𝑥 + 𝑥/√2)² = 16
⇒ 𝑥²/2 + 𝑥²(1 + 1/√2)² = 16
⇒ 𝑥²/2 + 𝑥²(3/2 + √2) = 16
⇒ 𝑥²/2 + 3𝑥²/2 + √2𝑥² = 16
⇒ 2𝑥² + √2𝑥² = 16
⇒ 𝑥²(2 + √2) = 16
⇒ 𝑥² = 16/(2 + √2)
⇒ 𝑥² = 16(2 - √2)/[(2 + √2)(2 - √2)]
⇒ 𝑥² = 16(2 - √2)/2
⇒ 𝑥² = 8(2 - √2)
∴ 𝑥 = 2√[2(2 - √2)]
As 𝑦 = 𝑥/√2
∴ 𝑦 = 2√(2 - √2)
Now, the area of △ABC, [△ABC] = ½.AC.BD = ½.𝑥.(𝑥 + 𝑦)
⇒ [△ABC] = ½(𝑥² + 𝑥𝑦)
½(𝑥² + 𝑥𝑦) = {8(2 - √2) + 2√[2(2 - √2)].[2√(2 - √2)]}/2
⇒ ½(𝑥² + 𝑥𝑦) = {8(2 - √2) + 4√2.[√(2 - √2)]²}/2
⇒ ½(𝑥² + 𝑥𝑦) = [16 - 8√2 + 4√2(2 - √2)]/2
⇒ ½(𝑥² + 𝑥𝑦) = (16 - 8√2 + 8√2 - 8)/2
⇒ ½(𝑥² + 𝑥𝑦) = 8/2
⇒ [△ABC] = ½(𝑥² + 𝑥𝑦) = 8/2 = 4
∴ [△ABC] = 4
This is honestly kind of similar to some of the problems shown on this channel. And on the second method, this shows how and why corresponding angles are congruent through right angled triangles. And that property establishes what the side measures irrespective of the acute angle. Bc that acute angle is contained in the full right triangle. I think tha I should practice BOTH methods when I can!
At 8:00, both ΔABD and ΔCAD are 22.5°-67.5°-90° triangles, for which the ratio of sides is known. The ratio (short)/(long) = √2 - 1, so AD/BD = CD/AD = √2 - 1. Let CD = x and AD = y, so BD = y + 4. Then, AD/BD = y/(x + 4) = √2 - 1 and x + 4 = y/(√2 - 1). CD/AD = x/y = √2 - 1 and x = y(√2 - 1). We substitute y(√2 - 1) for x in the first equation: y(√2 - 1) + 4 = y/(√2 - 1). Multiply both sides by (√2 - 1): y(√2 - 1)² + 4(√2 - 1) = y. Replace (√2 - 1)² by (√2 - 1) (√2 - 1) = 2 - 2√2 + 1 = 3 - 2√2: 3y - 2y√2 + 4√2 - 4 = y. Simplify: 2y(1 - √2) + 4(√2 - 1) = 0 and 2y(1 - √2) = -4(√2 - 1), 2y(1 - √2) = 4(1 - √2) and y = 2. Let BC be the base and AD the height, then area ΔABC = (1/2)(4)(2) = 4, as Math Booster also found.
*Outra Solução:*
Trace um segmento PB tal que PB=4 e o ângulo PBA seja 22,5. Após isso trace o segmento PA.
Ora, os triângulos BCA e BOA são congruentes, pois:
Os ângulos PBA=ABC=22,5°
AB é comum
PB=BC
Concluímos que os BAP=45°. Traçando o segmento CP, temos que os triângulos CBP e CAP são isósceles, além do mais, o triângulo CAP é retângulo, já que o ângulo CAP=90°.
Seja Q o ponto de intercessão dos segmentos CP e BA. Note que BA é a reta bissetriz com relação aos triângulos CBP e PAC, portanto, os segmentos BQ e AQ são as alturas respectivas desses triângulos.
[CBP]= 4×4 sin45°/2= 4√2, assim:
*[CBQ]=2√2*
Seja AP=AC=2L. Daí,
AQ=QP=QC=L√2, logo
[CQA]= L√2 × L√2/2= L^2, segue que:
*[ABC]=[CBQ] + [CQA]= L^2 + 2√2.*
Ora, como [CBQ]=2√2 então
QB ×L√2/2=2√2 => QB=4/L, por Pitágoras no ∆BQC:
16=16/L^2 + 2L^2 =>
2L^2 -16L +16=0, Pela fórmula de Bhaskara, temos:
L^2= 4+2√2 ou L^2= 4-2√2.
Repare que no ∆CBQ, o ângulo BCQ=67,5°> CBQ=22,5°, portanto, BQ> CQ => 4/L> L√2 => L^2
I added an angle =22.5 degrees to the angle B to form 45 degrees and extened AC to intersect in a point F. Then dropped a perpendicular on AB from the point C. Triangles AEC is iscoseles. Two equal trianglesCEB and CFB help to find CE. the rest is easy.
∠BCA = 180°-45°-22.5° = 112.5°. Draw a line CD where D is the point on AB where ∠DCA = 90°. As ∠DCA = 90°, ∠BCD = 112.5°-90° = 22.5°. As ∠BCD = ∠DBC, ∆CDB is an isosceles triangle and CD = DB = x.
As ∠ADC is an exterior angle to ∆CDB at D, ∠ADC = ∠BCD+∠DBC = 22.5°+22.5° = 45°. As ∠ADC = ∠CAD, ∆DCA is an isosceles triangle and CA = DC = x.
Draw DE, where E is the point on BC where DE is perpendicular to BC. As ∆CDB is isosceles, DE bisects ∆CDB and creates two new congruent right triangles ∆BED and ∆DEC. BE = EC = 2.
Extend BC to F and draw AF so that ∠CFA = 90°. As ∠BCA = 112.5°, ∠ACF = 180°-112.5° = 67.5°, and as ∠CFA = 90°, ∠FAC = 90°-67.5° = 22.5°. As ∠CFA = ∠BED, ∠FAC = ∠DBE, and AC = DB = x, then ∆CFA and ∆BED (and by extension ∆DEC) are congruent triangles. FA = BE = 2.
Triangle ∆ABC:
A = bh/2 = 4(2)/2 = 4 sq units
Draw a perpendicular from C to BA, sin 22,5 degrees=h:4, AB is divided into 3,7 and 1,53 the perpendicular is also 1,53 Hence A=4
Sine rule left the chat
|AB| from sine law
Area = 1/2*4*|AB|*sin(22.5)
AC=4 x sin 22.5/sin 45 = 2.1648 AD = AC x cos 22.5 =2 ABC = 4 x AD/2 = 4
*@ Math Booster* -- Thumbs-down to how the figure is drawn. In the figure, the measure of Angle B looks too close to the measure of Angle A, even though the measure
of Angle A is twice that of Angle B.
They don't look that close to me. If they were, then BC and AC would look close in length, which they do no.
Using a compass, I found angle B pretty close to 22.5 degrees, while angle A is about 40 degrees.
First'