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Point X is 3m from A, 4m from B and 5m from C. Point X is inside the triangle formed by ABC. If AB = BC and angle B is right angle. Find the length of side AB.

I have come up with the following figure and aware that this is an isosceles right triangle, unfortunately i'm not sure what to do next.

enter image description here

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  • $\begingroup$ You might try setting up coordinates for points A,B,C and letting X be $(x,y)$. Then write three distance equations for the given lengths. Of course initially there will be a parameter in the coordinates for A,B,C which will somehow have to be eliminated once you have the equations. $\endgroup$
    – coffeemath
    May 29, 2021 at 3:23
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    $\begingroup$ This should be what you need, using law of cosines. mathalino.com/reviewer/plane-geometry/… $\endgroup$
    – dfish
    May 29, 2021 at 3:30

3 Answers 3

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Assign coordinates to the 4 given points. B = $(0,0)$, C = $(L,0)$, A = $(0,L)$, X = $(x, y)$.

Apply Pythagoras to the three line segments.

  1. $$x^2 +y^2 = 16$$

  2. $$(L-x)^2 + y^2 = 25$$

  3. $$x^2 + (L-y)^2 = 9$$

Combining 1. and 2. gives $L^2 - 2Lx = 9$ and therefore $x = (L^2-9)/(2L)$

Combining 1. and 3. gives $L^2 - 2Ly = -7$ and therefore $y = (L^2 +7)/(2L)$

You can now substitute these expressions for $x$ and $y$ into equation 1 and solve for $L$. The result is $L = \sqrt{17 + \sqrt{224}}$, which is approximately 5.6539.

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  • $\begingroup$ The equation for L has a fourth order, a second order and a constant term. Set M = L^2. Then you have a quadratic equation in M which can be solved in the usual way. $\endgroup$
    – M. Wind
    May 29, 2021 at 16:08
  • $\begingroup$ Have you tried it? It's a simple biquadratic equation ... $\endgroup$ May 29, 2021 at 16:10
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enter image description here

Reflect point $X$ across the sides of $\triangle ABC$. Observe, $$\begin{align*} \angle DAF&=\angle DAX+\angle FAX=2(\angle BAX+\angle CAX)=90^{\circ}\;\implies DF=3\sqrt{2}\\ \angle DCE&=\angle DCX+\angle ECX=2(\angle ACX+\angle BCX)=90^{\circ}\;\implies DE=5\sqrt{2}\\ \angle FBE&=\angle FBX+\angle EBX=2(\angle ABX+\angle CBX)=180^{\circ}\implies F,\;B,\;E \;\;\text{are collinear.} \end{align*} $$ Using the cosine rule, in $\triangle EDF,$ $$\begin{align*} EF^2&=DE^2+DF^2-2\cdot DE\cdot DF\cdot\cos\angle EFD \\ 64&=18+50-60\cdot\cos\angle EFD \\ \cos \angle EFD&=\frac{1}{15} \implies \sin\angle EFD=\frac{4\sqrt{14}}{15} \end{align*} $$ Using the cosine rule, in $\triangle ADC,$ $$\begin{align*} AC^2 &=AD^2+CD^2-2\cdot AD\cdot CD\cdot \cos\angle ADC\\ &=9+25-30\cos(90^{\circ}+\angle EFD)\\ &=34+30\cdot\sin\angle EFD\\ &=34+8\sqrt{14} \\ \\ \therefore\; AB&=\sqrt{\frac{AC^2}{2}}=\boxed{\sqrt{17+4\sqrt{14}}\approx 5.6539} \end{align*}$$

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    $\begingroup$ Which interface did you use for drawing the figure? Pretty nice (+1) $\endgroup$
    – user907745
    May 29, 2021 at 6:00
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    $\begingroup$ GeoGebra $\endgroup$
    – Sathvik
    May 29, 2021 at 6:00
  • $\begingroup$ @SathvikAcharya : Please see my construction. I request you to comment on it. $\endgroup$
    – Narasimham
    May 29, 2021 at 9:00
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Comment

To construct the triangle in Geogebra I drew three concentric circles radii 3,4,5 for A,B,C respectively with X as center. Constructed a slidable right angle with 90 deg vertex at B. Drawn arc with radius 5.654 units from B to locate A and C as shown.

However got a different location of X inside the triangle with hypotenue AC almost 8 units long and X is almost on side AC. What may be my construction error?

enter image description here

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  • $\begingroup$ If you found points $A,C$ by constructing a circle centered at $B$ with radius 5.654 and finding the intersection points, how is the length of segment $BC=8$? $\endgroup$
    – Sathvik
    May 29, 2021 at 9:04
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    $\begingroup$ Typo corrected, else all ok. $\endgroup$
    – Narasimham
    May 29, 2021 at 15:57
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    $\begingroup$ Point $X$ is inside the triangle and all the conditions are perfectly satisfied. So it seems correct. (My figure is not drawn to scale) $\endgroup$
    – Sathvik
    May 29, 2021 at 16:00

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