Xuming's Geometry Synthetic Group - Shivam's proposal

  1. The circumcircle of the quadrilateral ABCDABCD has a radius 2. AC,BDAC, BD cut at EE such that AE=ECAE=EC. If AB=2AE,BD=23AB=\sqrt{2} AE, BD=2\sqrt {3} . Find the area of quadrilateral ABCDABCD.
  2. Triangle ABC is an isosceles triangle in which AB=A C. A circle is drawn passing through B and touching AC at its midpoint M. The circle cuts AB at P . Prove that BP=3AP.
  3. In an isosceles triangle the altitude drawn to the base is 23\frac{2}{3} times the radius of the circumcircle . Prove the base angle of the triangle is arccos23\text{arccos}\sqrt{\frac{2}{3}}.
  4. The sides of a right angled triangle are all integers . Two sides are prime that differ by 50.Find the smallest value of the third side.
  5. In triangle ABC , D is on BC such that BD=3DC. E is on AC such that 3AE=2EC. AD and BE cut F. Area of triangle AFE=4 and area of triangle BFD=30 . Find area of triangle ABC.
  6. In a circle AB is a diameter . AB is produced to P such that BP=Radius of the circle . PC is tangent to the circle . The tangent at B amd AC produced cut at E . Then describe triangle CDE.

Note by Shivam Jadhav
4 years, 1 month ago

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Q3:

Let the isosceles ABC\triangle ABC has AB=ACAB=AC, its base angle be θ\theta and the center and radius of the circumcircle be OO and rr respectively.

Then the extension of altitude APAP to BCBC passes through OO; AP=23rAP = \frac{2}{3}r, OP=13rOP = \frac{1}{3}r and then BP=OB2OP2=r2(13r)2=223BP = \sqrt{OB^2-OP^2} = \sqrt{r^2 - \left(\frac{1}{3}r\right)^2} = \frac{2 \sqrt{2}}{3}.

Therefore, tanθ=APBP=23r223r=12cosθ=23θ=cos1(23)\tan{\theta} = \dfrac{AP}{BP} = \dfrac{\frac{2}{3}r}{\frac{2\sqrt{2}}{3}r} = \dfrac{1}{\sqrt{2}} \quad \Rightarrow \cos \theta = \sqrt{\frac {2}{3}} \quad \Rightarrow \theta = \boxed{\cos^{-1} \left( \sqrt{\frac {2}{3}}\right)}

Chew-Seong Cheong - 4 years, 1 month ago

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Q5:

We note that CFE\triangle CFE has the same altitude as AFE\triangle AFE, therefore their areas are in the ratio of the base lengths. That is:

[CFE][AFE]=ECAE=32[CFE]=32×[AFE]=32×4=6\dfrac{[\triangle CFE]}{[\triangle AFE]} = \dfrac{EC}{AE} = \dfrac{3}{2} \quad \Rightarrow [\triangle CFE] = \dfrac{3}{2} \times [\triangle AFE] = \dfrac{3}{2} \times 4 = 6

Similarly, [CFD]=DCBD×[BFD]=13×30=10[\triangle CFD] = \dfrac{DC}{BD} \times [\triangle BFD] = \dfrac{1}{3} \times 30 = 10

And, [ABD]=BDDC×[ACD]=3×([AFE]+[CFE]+[CFD])=3×(4+6+10)=60[\triangle ABD] = \dfrac{BD}{DC} \times [\triangle ACD] = 3 \times ([\triangle AFE]+[\triangle CFE]+[\triangle CFD]) = 3 \times (4+6+10) = 60

Therefore, [ABC]=[ABD]+[ACD]=60+20=80[\triangle ABC] = [\triangle ABD] + [\triangle ACD] = 60 + 20 = \boxed{80}

Chew-Seong Cheong - 4 years, 1 month ago

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Q:1

Q:2

Ahmad Saad - 2 years, 6 months ago

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Answer 4 = 60

Saksham Rastogi - 4 years, 1 month ago

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