Germany Math Olympiad Challenge | Very Nice Geometry Problem
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- Опубліковано 1 бер 2024
- Germany Math Olympiad Problem | Circle and semicircle inside a triangle | Nice Geometry Challenge
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Solution with no trigonometry:
Locate Q, center of semicircle, and T, point where AC is tangent to semicircle.
Construct radius QT.
Triangle ABQ is congruent to triangle ATQ.
AT = 3
TC = 2
Angle QTC is a right angle because radius QT is perpendicular to tangent AC.
In right triangle QTC, TC =2 and QT = R
By the Pythagorean theorem, QC = SQRT(4+R^2)
BQ = BC - QC, so R = 4 - SQRT(4+R^2)
R = 3/2
Locate point M where AB is tangent to radius r.
Locate point P, center of circle.
Right triangle AMP is similar to right triangle ABQ
3 / (3/2) = AM/r
AM = 2r
MB = 3 - 2r
Construct PD, perpendicular to BC.
PD = MB = 3 - 2r
PQ = R+r = 3/2 + r
DQ = R-r = 3/2 - r
By the Pythagorean theorem, PD = SQRT [(3/2 +r) - (3/2 -r)]
PD = SQRT (6r)
SQRT (6r) = 3-2r
4r^2 - 18r + 9 = 0
Solve the quadratic equation for r, eliminating the answer where r > R.
r = (9 - 3 sqrt 5) / 4
r = 3/4 (3 - sqrt 5)
once the position of the center of the semicircle is known,
the line on which the second center point must be on, can be described, see line 140:
10 vdu5:for a=0 to 15:gcola:print a;:next a:vdu3:gcol8
20 @zoom%=@zoom%*1.4:nu=55
30 print "math booster-germany math olympiad challenge-very nice geometry problem"
40 dim x(2),y(2):l1=3:l2=4:sw=l1^2/(l1+l2)/10:ym1=0:r1=sw:yp=ym1
50 xg1=0:yg1=l1:xg2=l2:yg2=0:dx=xg2-xg1:dy=yg2-yg1:goto 90
60 xm1=r1:xp=xm1:zx=dx*(xp-xg1):zy=dy*(yp-yg1):k=(zx+zy)/(dx^2+dy^2)
70 xl=k*dx:xl=xl+xg1:yl=k*dy:yl=yl+yg1:lo=sqr((xp-xl)^2+(yp-yl)^2)
80 dg=r1/lo:dg=dg-1:return
90 gosub 60
100 dg1=dg:r11=r1:r1=r1+sw:r12=r1:gosub 60:if dg1*dg>0 then 100
110 r1=(r11+r12)/2:gosub 60:if dg1*dg>0 then r11=r1 else r12=r1
120 if abs(dg)>1E-10 then 110
130 l=sw:lm=sqr(l1^2+xm1^2):goto 160
140 xm2=xm1/lm*l:r2=xm2:dy=l/lm:ym2=l1*(1-dy):dgu1=(xm2-xm1)^2/l2^2:dgu2=(ym1-ym2)^2/l2^2
150 dgu3=(r1+r2)^2/l2^2:dg=dgu1+dgu2-dgu3:return
160 gosub 140
170 dg1=dg:lu1=l:l=l+sw:lu2=l:gosub 140:if dg1*dg>0 then 170
180 l=(lu1+lu2)/2:gosub 140:if dg1*dg>0 then lu1=l else lu2=l
190 if abs(dg)>1E-10 then 180
200 print l,"%",r2
210 x(0)=0:y(0)=0:x(1)=l2:y(1)=0:x(2)=0:y(2)=l1:masx=1200/l2:masy=850/l1
220 if masx 8 9 10 11 12 13 14 15
math booster-germany math olympiad challenge-very nice geometry problem
1.28115295% 0.572949017
run in bbc basic sdl and hit ctrl tab to copy from the results window
0 1 2 3 4 5 6 7
AC=5 radius of the larger semicircle : R
R/(4-R) = 3/5 5R = 12-3R 8R = 12 R = 3/2
√[3²+(3/2)²] = √(45/4) = 3√5/2 3/2 : 3 : 3√5/2 = 1 : 2 : √5
√5r+r+3/2 = 3√5/2 (√5+1)r = 3√5/2-3/2 = (3√5-3)/2
2r(√5+1) = 3√5-3 2r = (3√5-3)(√5-1)/4 r = (3√5-3)(√5-1)/8 = (18-6√5)/8 = (9-3√5)/4 = 3(3-√5)/4
Let O be the center of the semicircle and P be the center of the circle. LET D be the point of tangency between the semicircle and CA, and let E and F be the points of tangency between the circle and CA and AB respectively. Let R be the radius of the semicircle.
As AB = 3 and BC = 4, ∆ABC is a 3-4-5 Pythagorean triple triangle, And CA = 5. By two tangent theorem, DA = AB = 3, as DA and AB are tangent to the semicircle and meet at A. Similarly, EA and AF are similar, but the value is as-yet undetermined.
As DA = 3, CD = 5-3 = 2. As ∆CDO and ∆ABC share angles ∠C in common and are right triangles, ∆CDO and ∆ABC are similar.
OD/CD = AB/BC
R/2 = 3/4
R = 2(3/4) = 3/2
Draw OA. EA = AF and DA = AB, so DE = FB and ∆ABO and ∆ODA are congruent, since OB = OD = R. Similarly, ∆AFP and ∆PEA are congruent since PF = PE = r, therefore O, A, and P are colinear and OP = R+r.
OB² + AB² = OA²
R² + 3² = OA²
OA² = 9 + (3/2)² = 9 + 9/4
OA² = 45/4
OA = √45/4 = 3√5/2
PF/PA = OB/OA
r/(3√5/2-(3/2+r)) = (3/2)/(3√5/2)
(3√5/2)r = (3/2)(3√5/2-3/2-r)
3√5r/2 = 9√5/4 - 9/4 - 3r/2
3√5r/2 + 3r/2 = (9/4)(√5-1)
(3r/2)(√5+1) = (9/4)(√5-1)
r = (9/4)(√5-1)/(3/2)(√5+1)
r = (3/2)(√5-1)(√5-1)/(√5+1)(√5-1)
r = (3/2)(5-2√5+1)/(5-1)
r = (3/2)(6-2√5)/4
r = (9-3√5)/4 ≈ 0.573
Thank you for this video. Could you please share the reference of these Math Olympiad Questions?
Obviously, ⊿ABC is the famous 3 : 4 : 5 Pythagorean triangle. Therefore, the semicircle is half of the circle inscribed in an isosceles triangle with sides 6, 5 and 5.
Then we can use the general formula r = √((s − a) (s − b) (s − c) / s) where r is the radius of the incircle, s is the semiperimeter and abc are the sides of the triangle.
So, the radius of the semicricle is 3/2 or half of AB. Let's draw DE tangent to both circles and cutting PQ in T. Then AT = DE = R√5 − R and AD = AE = √5 (R√5 − R) / 2.
Then again, we can use the general formula, this time with a = 3/2 (√5 − 1), b = c = 3/4 (5 − √5) and s = 3. And we also find that r is 3/8 (√5 − 1)² or 3/4 (3 − √5) units.
Alternative to the latter half starting from 8:24 :
Make ST ⊥ AQ and ST intersects AB at T. Then MT=ST=BS, i.e., MB = 2 ST.
Since BQ is ½ AB, ST is ½ AS, i.e., ST = ¾ (√5 - 1).
Similarly, r is ½ of AM. But AM = 3 - MB = 3 - 2 ST.
So r = ½ (3 - 2 ST) = ½ (3 - 2 × ¾ (√5 - 1)) = ¾ (3 - √5 ).
With trigonometry or similar triangles i have got only ratio r/|AD| = 1/2 where D is tangent point
From comparing areas of triangle we can find radius of bigger circle
then we can consider auxiliary right triangle
This right trangle has side lengths (3-2r, 3/2 - r,r+3/2)
and from this I had Pythagorean theorem (3-2r)^2+(3/2 - r)^2 = (r+3/2)^2
I saw similar problem but r was given and area was searched
I had tried the method shown in the video, but for some odd reason I failed to recognize the length of AP. Then I went on to solve it in a much more difficult fashion, by drawing some auxiliary triangles and using twice the Law of Cosines. 😅
In fact law of cosines is much more easier fashion
Sir, once found AQ we can find r with similarity doing:
3/2 : r = 3/2√5 : (3/2√5 - 3/2 - r)
Agreed.
The large semicircle is not indicated as being exactly 1/2 of a full circle, so the center of the large semicircle is not known to lie on BC. The center of the large semicircle could lie on BC, but it could also be outside of the triangle ABC.
Also... It appears from the drawing, that the large semicircle is less than or equal to 1/2 of a full circle. But, there is no way to know from the information given, exactly what portion of a full circle it is. Due to inherent inaccuracies of relying on manually taking readings from a drawing, the large semicircle could even be slightly more than 1/2 of a full circle, placing the center of the large semicircle inside triangle ABC, near side BC.
This could have been avoided by indicating the location of the center of the large semicircle on the initial drawing.
Semicircle means 1/2 of a circle.
Quarter circle means 1/4 of a circle.
Why if you extend AP it will hit side BC exactly at point Q?
It doesn't have to hit side BC exactly at point Q, that it is the center of the semicircle.
Please let me explain why the Q is the center of lower circle.
Because it is semicircle so the centre will lie on the line BC.
Thanks, but How did you know the lower circle is a semicircle? Is there any comment or text in the question? I mean the lower circle is probably the part of circle but is it half? How can we know that?
Good point. Geometry is base of proof.